Magnetic Field Due To A Current

An electric current flows in a wire from north to south. So the direction of the magnetic field associated with it can be predicted by Maxwell’s Right hand Cork-Screw rule.

So we can conclude that,

(a) The magnetic field to the east of the wire will be directed vertically upwards.

(b) The magnetic field to the west of the wire will be directed vertically downwards.

(c) The magnetic field vertically above the wire will be directed towards the west.

(d) The magnetic field vertically below the wire will be directed towards the east.

Magnetic field at a point d distance apart from a long straight current carrying wire is given by from Ampere’s circuital law. Where μ₀ is the magnetic permeability of free space, i denotes the current through the wire and d denotes the distance from the wire, of the point where magnetic field is to be calculated.

Now we know that, speed of light, (ϵ₀ denotes the electric permittivity of free space)

c² = μ₀ ϵ₀

Thus we get magnetic field to be

We can predict the direction of the magnetic field by Maxwell’s Right hand Cork-Screw rule. So we put the thumb of our right hands pointing towards the direction of current and now the direction the other fingers curl toward will be direction of the magnetic field.

So, we get the field at the center is moving away from me if the current is going in the clockwise direction.

The contribution to the magnetic field B by currents outside the Amperian loop turns out to be zero because if we draw a straight line on any fixed point on the loop from the current vector(the direction vector of current), let line A and draw another line from the current vector to a movable point on the loop, let line B, and move that point in one direction along the loop until the full loop is traversed then we can see the net positive angle travelled by line B with respect to line A is equal to the net negative value travelled by the same. So the total angle travelled is zero. The line integral has a finite value for a current that is within the loop as it traverses an angle of 2ϖ. So, we can say that Ampere’s law takes into account every current be it inside the Amperian loop/curve or outside but as for outside currents, the integral is zero, so only the enclosed current is taken into account for calculation but actually takes into consideration the contribution of all the currents.

we consider the circle to be the Amperian loop. Point B is the fixed point C is the moving point. At some point of time C coincides with C’ where ∠BAC = -∠BAC’. We will get a point like this for every arbitrary point on the circle, so total angle for all the circle will be zero as stated before.

Magnetic field on the axis of a circular current carrying loop is given by,

where a denotes the radius of the loop and x is the distance of the point on the axis at which the magnetic field is to be calculated. i is the current flowing in the loop. μ₀ is the magnetic permeability of free space.

And B = μ₀ni (where μ₀ is the magnetic permeability of free space, n is the number of turns of the wire per unit length of the solenoid, and i is current in the wire) holds true for a very long solenoid where at the terminal points x >> a. We can say that if we keep adding more loops at the ends, the total number of loops increase but with that length also increases so the turns per unit length n is unvaried. We can also think of it this way,

The value of B for loops added to the ends of the solenoid will have dropped to almost zero as for the condition x >> a,

This can be shown mathematically by taking the limit of B for a circular loop. So, let us take x to be tending to infinity and a to have a finite value.

(Dividing both sides by x³)

So if we add more loops at the ends of a very long solenoid, the magnetic field due to each one of them will be zero at the centre of the solenoid thus won’t have considerable effect.

Ampere’s law holds true for any closed loop of magnetic field containing constant electric current. The left hand side of Ampere’s circuital law

Where

μ₀ is the magnetic permeability of free space,

i is the current enclosed by the loop and B is the magnetic field

The above equation deals with a closed line integral along the path of the magnetic field. The line integral will not have a finite value if the loop is not closed. So, Ampere’s law will not hold for a loop that does not enclose the wire but it will hold for any shape of the loop even if it’s not circular.

No, there won’t be any magnetic force acting on the wire in this case because, if the ring starts rotating, it will create a magnetic field which is parallel or antiparallel to the direction of current or the length vector of the wire at the center of the ring. i.e. angle between the is zero or ϖ respectively. According to F = ILBsinϴ

for ϴ = 0 or ϖ,

F= 0.

Here I is the current, L is the length of the wire and B is the magnitude of the magnetic field.

So, there will be no force on the wire if the ring starts rotating about the wire.

Using the relation for force acting on a conductor

Where

I is the current in the wire,

L is the length vector of the wire

B is the external electric field.

Thus, if the wires are kept perpendicular, the angle between them will be 90 and thus the angle between the movable wire and the magnetic field due to the fixed wire is 0( from Maxwell’s Cork-Screw rule). Thus the force is the minimum and it is in an unstable equilibrium. Thus the will rotate and make itself parallel to the fixed wire and attain the maximum attractive force and thus a stable equilibrium.

The magnetic field is not as strong as electric field and the magnitude of magnetic field due to a moving charge only equals that of its electric field only when the velocity is equal to the speed of light. So for two protons moving in the same direction, Force due to electric field repulsion will exceed that due to magnetic field attraction. So the proton beams moving in the same direction repel each other.

But two parallel current carrying wire attract due to magnetic attraction as the electric field is inside the conductor and is shielded by it and can’t be felt outside. The magnetic attraction can be further reinforced by Fleming’s left rule by considering the magnetic field produced by one wire and electric field in the other wire in the direction of current.

We can always find the magnetic field due to a long, straight current carrying wire along any loop which only contains that wire by using Ampere’s circuital law. The law only takes into account the current bound in the loop irrespective of the source. If we take the whole circuit along with the battery within that Amperian loop, then the total current contained in that loop will be zero so the resultant magnetic field will be zero too even if the wire itself produces a non zero magnetic field.

If we choose an Amperian loop containing both the wires, then the magnetic field along that loop will turn out to be zero as the wires carry currents in the opposite direction and the total current turns out to be lesser than the maximum of that of the two wires. And if the magnitudes of currents are the same then the total current in that loop turns out to be zero. If they are kept very close together, this will hold for even very small Amperian loops. Thus twisting two wires, which carry currents in the opposite directions, minimizes the magnetic field produced and can even cancel it out.

denotes current going perpendicularly inside the screen.

denotes current coming perpendicularly out of the screen.

If this both currents are equal in magnitude then the total current enclosed by the Amperian Loop C will be zero.

The displacement of the wire due to magnetic forces happens in the direction of the distance vector between the two wires. The magnetic field due to the wires is perpendicular to the wire and the distance vector between the wires so work done is by the field is zero as usual. As the wires move towards each other there is a change in magnetic field and according to Lenz law there will be an induced current in the wires which will be directed opposite to the current initially carried by them. As these electrons cannot come out of the wire, they hit the edges of the wire starts moving and gains kinetic energy.

With the right hand cork screw we can get the direction of the magnetic field due to the vertical current carrying wire. And we find it to be directed downwards. Now applying Fleming’s Left hand rule, we can find the direction of force as we put the direction of the electron beam as the opposite direction of current. We can find that force is pointed upwards. So the electron beam will be deflected upwards.

There won’t be any magnetic force acting on the wire in this case because, according the Cork-Screw rule, the magnetic field due to the straight wire will be parallel or anti parallel to the direction of current in the loop, thus creating an angle of 0 or ϖ with the direction of current.

According to F = ILBsinϴ

Where

I is the current,

L is the length of the wire

B is the magnitude of the magnetic field.

for ϴ = 0 or ϖ, F= 0.

We can consider the beams to be current carrying wires. The proton beam is going from north to south so the current for it will be directed from north to south. And the electron beam is moving from South to north so the current will be directed opposite to the direction of the beam, i.e. from north to south. So we get the case of two current carrying wires carrying current in the same direction and they attract each other. So the electron beam will be deflected towards the proton beam.

If the current is directed towards north at the topmost point of that circular loop that means, from point A’s perspective, current is in clockwise direction and will cause a magnetic field which is moving inwards i.e. to the west at point A and subsequently to the west at point B.

The closest side of the loop to the straight wire carries a current in the same direction as the straight wire. So there will be an attractive force. And for the opposite side of the loop the current is moving in the opposite direction to that of the straight wire, so there will be a repulsive force. As the closest side shows attractive force and the farthest side shows repulsive force, the attractive force will be dominant. To explain it mathematically, we introduce the expression for force per unit length on each of the wires.

Where

F denote the force per unit length,

I₁, I₂ denote the currents in the two wires,

d is the distance between them

μ₀ is the magnetic permeability of free space.

The current in the loop is of same magnitude everywhere. The parallel sides of the loop to the long straight wire will encounter all the force. As the side closer to the straight wire carries a current in the same direction, I₁I₂ will be positive thus it will encounter attractive force and the opposite side will encounter repulsive force as the, I₁I₂ is negative as the currents are in opposite directions. We can see that the force is inversely proportional to the distance. As the wire carrying current in the same direction is closer than the wire carrying current in the opposite direction, the attractive force will be greater than the repulsive force. The other two sides will experience forces of equal amplitude but opposite directions and thus will cancel each other out. Thus the loop will move towards the wire.

If a charged particle is moving along a magnetic field line, that means it is moving either parallel or anti parallel to the direction of magnetic field. i.e. the direction of velocity of the particle is either making an angle of 0 or ϖ. According to F = qvBsinϴ (where q is the charge of the particle, v is the velocity of it and B is the magnitude of the external Magnetic field and ϴ is the angle between the direction of magnetic field and the velocity) for ϴ = 0 or ϖ, F= 0.

Under any circumstances, a charge particle will produce an electric field. And if we imagine that the moving charge particle is analogous to an electric current then it is clear to us that it, most certainly, will produce a magnetic field as well.

If a charged particle is projected with velocity v on a plane perpendicular to a uniform magnetic field B, then the particle will travel in a circle. Whose radius is where m and q are the mass and the charge of the particle respectively.

So the area enclosed by the path will be A = ϖR² = = = kinetic energy

Thus, the area will be proportional to the kinetic energy of the particle.

let the charge of both of them be q and mass be m_x and m_y respectively. If they are driven through the same potential V then,

Thus, v_x and v_y (their respective velocities after leaving the potential) will be and

So from we get, (all the other terms are constant for both the particles)

So, according to the question

Thus,

The third wire is carrying a current 20A antiparallel to the previously set 20A current wire. So these wires will repel each other. As the 40A wire is carrying a current in the opposite direction to the previously set 20A wire, the newly set 20A wire will be having the same current direction to it. So they will attract. So the newly set wire in the midway will move towards the 40A wire.

The magnetic field caused by a long wire is given by where i is the current in the wire, d is the distance at which B is to be calculated, μ₀ is the magnetic permeability of free space. Let the distance between the two wires be 2d.

Then, when the currents are parallel, one’s magnetic field will negate the other’s because at the point midway between the two wires, the magnetic fields generated by them will be directed in the opposite directions and it can be verified by the right hand Cork-Screw rule.

So the total B will be

which is given to be 10μT and when the direction of i₂ is reversed then, is given to be 30μT. So, solving these, we get,

(Componendo- Dividendo rule)

We know, drift velocity of electron is

vd = i/(nAe)

Where

i is the current in the wire,

n is the electron number density of the wire a,

A is the area of cross section of the wire,

e is the charge of the electron.

We can see that the trolley is moving with drift velocity but current travels in a conductor with a speed comparable to that of light in free space. So the velocity of the trolley is very low compared to the velocity at which the current travels. So there will be no significant change in the value of B and it will be .

According to Biot-Savart law:
The magnetic field due to a current element is given as
Where,

i is the current,

is the length element vector,

is the vector joining the required point and the length element,

μ₀ is the permeability of vacuum and μ₀ = 4π × 10^-7 TmA^-1.
Also, by vector properties: A× B = -(B× A)
Thus,
Thus, options (A) and (B) are the correct options

(A)Dimension for electric field E is: F/q = [MLT^-3A^-1]
Force due to magnetic field is given as: F=qvB.
Where q is the charge, v is the velocity of charge and B is the magnetic field.
Hence

The dimensions of charge, q is [AT]

The dimension of the velocity, V is [LT-1]

Putting the value in the above formula, we get

Dimensions of B =

∴ Dimensions of B = [MT^-2A^-1]
Thus,

(B)
We know that Where

μ₀ is permeability of free space,

ϵ₀ is the permittivity of free space and

c is the speed of light.
Thus, y = c, c is the speed and unit of speed is m/s
Dimensions of y = [LT^-1]
(C)

Reactance,
Where

C is the capacitance

R is the resistance

l is the length of the wire.

In a CR circuit the product CR is the time constant and unit is seconds.
Thus, Dimensions of z = l/CR = m/s= [LT^-1]
Hence x, y and z have same dimensions.
Thus, options (A),(B) and (C) are the correct options.

When a current carrying wire is along z-axis, the magnetic field can be found using right hand thumb rule.

Here X represents the field is going in and O represents field is coming out.
First case, if two points in the XY plain are above the wire and at same distance from z-axis, then both will lie in the X region(field going in), thus have same direction of magnetic field and magnitudes of magnetic field would be same.
Second case , if the two points in XY plane are at same distance from z axis but one point is above the wire and one is below it, then the direction of the electric field would be opposite for both the points. One point having X direction (going in) and other having O direction (coming out).
First two cases satisfy options (B),(C) and (D).
Thus, options (B),(C) and (D) are the correct options.

According to Ampere’s Law:
Where,

dl is the current element,

B is the magnetic field,

μ₀ is the permeability of free space and

i is the current flowing.

Thus, at the cross-section the formula becomes,
2πR is the circumference of the wire and R is the radius. We get,
Now, at the axis of the wire. R=0, and so no area to integrate and hence zero current is enclosed. Thus magnitude of magnetic field is minimum at axis of the wire.
At the surface of the wire, R= some minimum value.
As R increases , magnitude of magnetic field decreases.
Hence, B will be maximum at the surface of the wire.
Thus, options (B) and (C) are correct options.

By Ampere’s Law ,
Here, dl is the current element, B is the magnetic field, μ₀ is the permeability of free space and i is the current flowing.
We know that , magnetic field due to a circular wire is given
R is the radius of the wire.
Consider a Hollow tube as the series of circular wires.
Inside the tube at a point, r<R
Current inside the hollow tube is zero: i_inside = 0

Hence , the magnetic field is constant inside the hollow tube.
Now, at the axis, R=0.

Therefore, zero current is enclosed.
Hence, B=0.
Thus, options (B) and (C) are the correct options.

According to Ampere’s Law:
Where,

B is the magnetic field,

dl is the current element,

μ₀ is the permeability of free space

i is the current flowing in the conductor

Magnetic field outside the cable: B = B_I+B_O
B_I and B_O are the magnetic field of the inner conductor and outer cable respectively.
Due to wire,
R is the radius of the wire.


Thus, option (A) is the correct option.

Magnetic field inside the inner conductor:

∮B. dl =μ_o I (I=0(inside the wire))

Magnetic field between the two conductors and inside the outer conductor.

∮B. dl =μ_o I

On integrating the length element over the complete loop, we get dl= 2π r

B.2π r=μ_o I

On solving the above equation for B, we get

Thus, option A and B is the correct option.

Using Ampere’s law, the electric field due to a current carrying conductor of cross section areas is:
magnetic field,
Where,

dl is the current element,

μ₀ is the permeability of free space,

i is the current flowing

R is the radius of the conductor.
At the axis,

R=0 and so no area to enclose current.

Hence, B=0 at the axis of the conductor.

Electric field in the vicinity of the conductor is zero as a minimum distance between the point and the conductor is required to have electric field at that point.
Thus, options (B) and (C) are correct options.

Given:

Formula used:
Lorentz force:
Where, is the magnetic force vector, q is the charge, is the velocity vector and is the magnetic field vector.

Magnetic field can be written as:
Unit of force is Newton: N
Charge: q = It
Unit of current : I = Amperes= A
Unit of time : t = seconds = s
Unit of charge : q = As
Unit of velocity : v = m/s
Thus, unit of Magnetic field: Therefore, unit of Magnetic field : B = N m^-1A^-1.
Now,
Ampere’s Law:
Where,

μ₀ is the permeability of free space,

I is the current flowing

r is the radius of the current carrying wire.
SI Unit of B = N mA^-1
SI Unit of I = A
SI Unit of r = m


Thus, unit of permeability constant μ₀ is N A^-2.
Hence proved that unit of Magnetic field is N m^-1A^-1 and unit of permeability constant μ₀ is N A^-2.

Given:
Current : I = 10 A

Point is located along x-axis at 1 m.
Formula used:
Using Ampere’s Law for a current carrying straight wire:
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point.
Substituting the values,

∴B = 2×10^-6 T
Hence, magnetic field at a point (1m, 0, 0) is 2× 10^-6 T.

Given:
Diameter of the copper wire: d = 1.6mm = 1.6× 10^-3 m
Current through the wire: I = 20 A
Formula used:
By Ampere’s law for current carrying wire of cross section area,
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 r is the radius of the wire.

Radius of the wire : r = d/2 = 8× 10^-3 m
Substituting we get,

∴ B = 5× 10^-3 T
Hence, maximum magnitude of magnetic field due to a current of 20 A is 5× 10^-3 T.

Given:
Current through the transmission wire: I = 100 A
Distance between wire and the road : d = 8 m
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space

μ₀= 4π × 10^-7 T mA^-1

d is the distance between the current carrying wire and the required point.
Substituting we get,

B = 2.5× 10^-6 T
Hence, magnitude of magnetic field at a point on the road due to current carrying transmission wire is 2.5× 10^-6 T.

Given :
Current in the wire : i = 1.0 A
Magnitude of Horizontal magnetic field: B_o = 1.0 × 10^–5 T
Distance between points P,Q and the wire :
d = 2.0 cm = 0.02 m

Right hand rule is used to determine the direction of the magnetic field due to the current carrying wire. Cross determines the field going into the plane and dot determines the field coming out of the field.
B_w is the magnetic field due to wire.
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space, μ₀= 4π × 10^-7 T mA^-1

d is the distance between the current carrying wire and the required point.

Magnetic field at point P due to wire,


Resultant magnetic field at P
B_P = B_o +B_w
∴ B_P = 1× 10^-5 + 1× 10^-5
∴ B_P = 2 × 10^-5 T
Magnetic field at point Q due to wire,

∴ B_w = - 1 × 10^-5 T
The negative sign is because the field at Q is opposite to the direction of horizontal field and at P.
Resultant magnetic field at P
B_Q = B_o +B_w
∴ B_Q = 1× 10^-5 - 1× 10^-5
∴ B_Q = 0
Hence, resultant magnetic field at P is 2 × 10^-5 T and at Q is 0.

Given:
Radius of the wire : r
Current through the wire : i
Uniform magnetic field pointing upward : B

Formula used:
Magnetic field due to a current carrying wire is

Magnitude of magnetic field due to wire,

Where, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1.

r is the radius of the wire.
(a)
Maximum net magnetic field would be above the wire at P.
Considering direction of the current as shown in the diagram, right hand rule will give us direction of the magnetic field due to wire.
thus,
B_net = B+B_w
∴ B_net = B +
(b)
Similarly, the field at Q would be in opposite direction to the P.
Minimum magnitude of the magnetic field will be a Q
B_net = B-B_w
∴ B_net = B -
Hence, magnitude of magnetic field will be maximum at points above at wire and minimum at points below the wire only if the direction of the current is as shown in the figure.

Given:
Current in the wire : I = 30 A
External Uniform magnetic field: B_O = 4.0 ×10^–4 T
Distance between the point and the wire : d = 2 cm = 0.02 m

Here, B is the magnetic field due to wire. B_O is parallel to the current in the wire. Using Right hand rule we can see that direction of magnetic field at the desired point is going into the plane (cross). This direction due to wire is perpendicular to the External magnetic field.
Formula used:
By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point.
Substituting we get,

∴ B = 3× 10^-4 T
As Magnetic field due to wire is perpendicular to the external magnetic field, the resultant net magnetic field would be:


∴ B_net = 5 × 10^-4 T
Hence, the magnitude of the resultant magnetic field at a point 2.0 cm away from the wire is 5 × 10^-4 T.

Given:
Current in the wire : I = 10 A
Magnitude of Horizontal magnetic field : B_O = 2.0 × 10^–3 T

For the resultant magnetic field to be zero, we need to have a field in opposite direction to that of the existing field.
Hence, magnetic field B due to wire in the direction from north to south as shown in the diagram..
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 d is the distance between the current carrying wire and the required point.
To get zero resultant,
B=B_O


∴ d = 0.001 m=1 mm
Hence, the point should be place at a distance of 0.001 m from the wire in west direction

Given:
Magnitude of current in both the wires: i = 10 A
Distance between 2 wires : d = 4cm = 0.04 m

P and Q are the two wires having opposite direction of currents. From the diagram, P has current into the plane and Q has current coming out of the plane.
By right hand rule, P will have field coming up at A₁ and going down at A₂ and A₃ and tangent to A₄ as shown by the red arrow.
Similarly, Q will have field going down as A_1,A₂ and A₃ aand tangent to A₄ as shown by blue arrow.
Distance between, A₁,A₂ and A₃ is 2 cm each.
Formula used:

By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point.
At A₁ :
Net magnetic field due to P and Q would be subtraction of individual fields at A₁.
PA₁ :d₁ = 0.02 m
QA₁ : d₂ = 0.06m
B_net = B_P – B_Q
Where, B_P is the magnetic field due to wire P and B_Q is the magnetic field due to wire Q.


∴ B_net = 6.66 × 10^-5 T
Hence, magnetic field at A₁ is 6.66 × 10^-5 T.
At A₂
Effect of magnetic field at A₂ will be sum of magnetic field due to P and Q as both are in same direction as shown.
PA₂: d₁ = 0.01 m
QA₃: d₂ = 0.03 m
B_net = B_P + B_Q
Where, B_P is the magnetic field due to wire P and B_Q is the magnetic field due to wire Q.



∴ B_net = 2.66 × 10^-4 T
Hence , magnetic field at A₂ is 2.66 × 10^-4 T
At A₃
Effect of magnetic field at A₃ will be sum of magnetic field due to P and Q as both are in same direction as shown.
PA₃: d₁ = 0.02 m
QA₃ : d₂ = 0.02 m
B_net = B_P + B_Q
Where, B_P is the magnetic field due to wire P and B_Q is the magnetic field due to wire Q.



∴ B_net = 2× 10^-4 T
Hence, magnetic field at A₃ is 2× 10^-4 T
At A₄:
The resultant magnetic field at A₄ would be
From Pythagoras theorem,
(PA₄)² = (PA₃)² + (A₃A₄)²
∴ (PA₄)² = 0.02² + 0.02²
∴ PA₄ = 0.028 m = d₁
Now,

∴B_P = 7.14 × 10^-5 T
Similarly,
QA₄ = 0.028 m = d₂
Thus,


∴ B_Q = 7.14 × 10^-5 T
Substituting to get net magnetic field at A₄,

∴ B_net = 1× 10^-4 T
Hence magnetic field at A₄ is 1× 10^-4 T.

Given:
Current in the equal wires: i= 10 A
Distance between the two wires: a = 2 cm = 0.02 m
Distance between the wire and the point : d = 2 cm = 0.02 m

From the diagram l₁ and l₂ are the wires coming out of the plane.
These two wires and point P form an equilateral triangle of side 0.02 m.
By right hand rule, Magnetic field at P due to l₁ is shown by arrow B₁ tangent to the field’s path and Magnetic field at P due to l₂ is shown by B₂ tangent to the field’s path.
By geometry we can calculate the angle between B₁ and B₂ which turns out to be 60° .
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space, μ₀= 4π × 10^-7 T mA^-1

d is the distance between the current carrying wire and the required point.
Magnetic field due to wire l₁:

∴ B₁ = 1 × 10^-4 T
Magnetic Field due to l₂ :
∴ B₂ = 1 × 10^-4 T
B₁=B₂ as same magnitude of current is flowing through both the wires and point P is located at same distance from each wires.
Now, angle between and : θ = 60°
Resultant of two vectors is given as
B_R is the resultant magnetic field at P.

∴ B_R = 1.73 × 10^-4 T
Hence, resultant magnetic field at a point 2 cm away from due to two current carrying wires is 1.73 × 10^-4 T.

Given:
Current in both the wires : I = 5 A.

Wires 1 and 2 generate magnetic fields B₁ and B₂ respectively.
By right hand rule, direction of magnetic field in each quadrant due to wires 1 and is shown by X and O.
X : Field is going into the plane
O : Field is coming out of the plane.
Formula used:
By Ampere’s Law for a current carrying wire is
Where,

B is the magnitude of magnetic field,

μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 d is the distance between the current carrying wire and the required point.
(a) At (1m,1m)
As we can see from the diagram at (1,1) the magnetic fields due to wire 1 and 2 are in opposite direction. (X and O)
Hence, net magnetic field would be zero.
(b) At (-1m , 1m)
Here, magnetic field due to wire 1 and 2 would add up as direction of magnetic field is same. ( O and O)
Thus B = B_X + B_Y
B_X is the magnetic field due to wire on x axis
B_Y is the magnetic field due to wire on y axis

∴ B = 2× 10^-6 T
This B would be along z-Axis.
(c) At (-1m,-1m)
As we can see from the diagram at (-1,-1) the magnetic fields due to wire 1 and 2 are in opposite direction. (O and X)
Hence, net magnetic field would be zero.
(d) At (1m,-1m)
Here, magnetic field due to wire 1 and 2 would add up as direction of magnetic field is same. ( X and X)
Thus B = B_X + B_Y
B_X is the magnetic field due to wire on x axis
B_Y is the magnetic field due to wire on y axis

∴ B = 2× 10^-6 T
This B would be along negative z-axis.

Given:
Current in the 4 wires: i = 5.0 A
Side of the square : d = 5 cm = 0.05 m

In the diagram above, l₁ and l₂ represents vertical wires. l₃ and l₄ represent horizontal wires.
Arrows represent the direction of current in the wire. We can determine the direction of magnetic field due to any of the wire using Right hand rule.
Formula used:
By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point.
(a)
At point P, if we use Right hand rule for every wire.
Field due to l₁ would be into the plane and field due to l₂ would come out of the plane. Thus canceling out each other.
Similarly, field due to l₃ would go into the plane and field due to l₄ would come out, canceling each other.
Thus, net field at P is zero.
(b)
At Q_1, By geometry:
Distance between wire l₁ and Q₁ : d₁ = 0.025 m
Distance between wire l₂ and Q₁ : d₂ = 0.075 m
Distance between wire l₃ and Q₁ : d₃ = 0.025 m
Distance between wire l₄ and Q₁ : d₃ = 0.075 m
Net magnetic field at Q₁ due to wires l_1,l_2,l_3,l₄ is
B_Q1 = B_l1+B_l2+B_l3+B_l4


∴ B_Q1 = 1.06 × 10^-4 T
Hence, magnetic field at Q₁ is 1.06 × 10^-4 T and in upward direction.
At Q₂, Similarly by geometry
Distance between wire l₁ and Q₂ : d₁ = 0.075 m
Distance between wire l₂ and Q₂ : d₂ = 0.025 m
Distance between wire l₃ and Q₂ : d₃ = 0.025 m
Distance between wire l₄ and Q₂ : d₃ = 0.075 m
Now, Net magnetic field at Q₂ due to wires l_1,l_2,l_3,l₄ is
B_Q2 = B_l1+B_l2+B_l3+B_l4


∴ B_Q2 = 0
Hence, magnetic field at Q₂ is zero.
Negative sign for B_l3 and B_l4 as direction of magnetic field due to l₁(going in) is opposite to l₄(coming out) and l₂(going in) is opposite to l₃9coming out).
At Q₃, By geometry
Distance between wire l₁ and Q₃ : d₁ = 0.075 m
Distance between wire l₂ and Q₃ : d₂ = 0.025 m
Distance between wire l₃ and Q₃ : d₃ = 0.075 m
Distance between wire l₄ and Q₃ : d₃ = 0.025 m
Net magnetic field at Q₃ due to wires l_1,l_2,l_3,l₄ is
B_Q3 = B_l1+B_l2+B_l3+B_l4


∴ B_Q3 = 1.06 × 10^-4 T
Hence, magnetic field at Q₃ is 1.06 × 10^-4 T and in downward direction.
At Q₄,
Magnetic field at Q₄ would be same as that of Q₂ because at Q₄, l₁ and l₄ have opposite fields, l₂ and l₃ have opposite fields. Thus canceling out each other and B_Q4 is zero.
Hence, Magnetic field at Q₁ and Q₃ have same magnitude of
1.06 × 10^-4 T but in opposite direction and magnetic field at Q₂ and Q₄ is zero.

Given:
Current through the wire : i
radius of the circle formed: d

Let W₁ and W₂ be the terms for wire 1 and wire 2.
Arrow shows the direction of the current in the wire.
By right hand rule we can determine the direction of the magnetic field due to any of the two wire at any point.
Formula used:
By Ampere’s Law for a current carrying wire is
Here, B is the magnitude of magnetic field, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1 and d is the distance between the current carrying wire and the required point. In this case, d is the radius of the circle.

(1) At point P,
Magnetic field at P due to W₁ is zero as P is on the axis of W₁.
Magnetic field at P due to W₂ is
Hence, net magnetic field at P due to both the wires is
Direction of B_P is perpendicular to the plane and in outward direction.
(2) At point Q,
Magnetic field at Q due to W₁,
Magnetic field at Q due to W₂ is zero as Q is not in the effective zone of W₂.
Hence, net magnetic field at Q due to both the wires is
Direction of B_Q is perpendicular to the plane and inward direction
(3) At point R,
Magnetic field at R due to W₁ is zero as R is not in the effective zone pf W₁.
Magnetic field at R due to W₂ is
Hence, net magnetic field at R due to both the wires is
Direction of B_R is perpendicular to the plane and in inward direction.
(4) At point S,
Magnetic field at S due to W₂ is zero as S is on the axis of W₂.
Magnetic field at S due to W₁ is
Hence, net magnetic field at S due to both the wires is
Direction of B_P is perpendicular to the plane and in outward direction.

Given:
Length of the piece: x
Current in the piece : i
Distance between midpoint of piece and P : d

P is on the perpendicular bisector of AB meeting at O.
Formula used:
By Biot-Savart Law:
Here, dB is the magnitude of magnetic field element, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1,dl is the length element, r is the distance between the current carrying wire and the required point.
We know that, magnetic field at a point on the perpendicular bisector is
Here, a is the length of the wire. In this case a=x
d is the distance between point P and the midpoint of AB.
For d≫x
x² can be neglected.



∴ B α 1/d²
Hence for d≫x magnetic field is inversely proportional to the square of the distance d.
For d≪x
d² can be neglected.

∴ B α 1/d
Hence, when d≪x the magnetic field is inversely proportional to the distance d.

Given:
Length of the wire piece: x = 10 cm = 0.1 m
Current in the wire piece: i= 10 A

Formula used:
By Biot-Savart Law:
Here, dB is the magnitude of magnetic field element, μ₀ is the permeability of free space and μ₀= 4π × 10^-7 T mA^-1,dl is the length element, r is the distance between the current carrying wire and the required point.
We know that, magnetic field at a point on the perpendicular bisector is
Here, x is the length of the wire.
d is the distance between point C and the midpoint O.
From Pythagoras theorem,
(BC)² = (OB)²+(OC)²
∴ (OC)² = (BC)² - (OB)²
∴ (OC)² = (0.1)² – (0.05)²
∴ OC = (7.5× 10^-3)^1/2
∴ OC = 0.086 m = d
Substituting for B we get,

∴ B = 2.32× 10^-5 × 0.502
∴ B = 1.16 × 10^-5 T
Hence the magnitude of the magnetic field due to the piece at a point which makes an equilateral triangle with the ends of the piece is 1.16 × 10^-5 T.

Magnitude of current = i

Distance of point P from wire = d

The magnetic field induced in the wire is defined as

Also, the magnetic field due to a section of length l on the perpendicular bisector is given by

Neglecting (as it is very small), we get

now,

We are given that,

⇒

substituted the value of in the above equation we get,

⇒

⇒

⇒

Resistance of wire ADC is given twice that of wire ABC. hence current flow through ADC is half of ABC

i.e.,

--------------------(1)

also ---------------(2)

Equating equations (1) and (2), we get

and

Hence, magnetic field at centre O due to wire AB and BC

and magnetic field at centre o due to wire AD and DC

Also , so

Hence net magnetic field at centre O

magnetic field due to AB

magnetic field due to DC

The magnetic field due to AD and BC are equal and opposite hence nullify each other

Hence, net magnetic field

Magnetic vector B due to BC and due to AD at PT equal and opposite.

Hence net vector B = 0

Similarly, due to AB and CD at P = 0

∴ The net vector B at the centre of the square loop = 0

Let current 2I flow through the circuit.
Since the wire is uniform, the current will enter the loop at point A and divide equally between the two arms AB and AC (as shown in the figure)

Now,
Magnetic field induced at point P due to wire AB and wire AC = B (say)
(Perpendicular to the plane in outward direction)

Magnetic field at P due to wire BD and DC = B' (say)
(Perpendicular to the plane in outward direction)

∴ Net magnetic field at = 0

(a)

Let M be the midpoint of equilateral ΔABC

∴

current is given to be i

Hence, in ΔAOB

The angles made by points B and C with centre M are respectively,

Hence separation of the point from the wire is given by,

Thus, the magnetic field induced due to current in wire BC is-

Now,
Net magnetic field at point M is given as Magnetic field due to wire BC + Magnetic field due to wire CA + Magnetic field due to wire AB
Since all wires are equal,

It is perpendicular to the plane in outward direction if current is anticlockwise and in an inward direction if the current is clockwise.

(b)

The angles made by B and C with centre M are

Distance between point from the wire,

Thus, the induced magnetic field due to electric current in wire BC is given by

Since all wires are equal,
Net magnetic field at point M = 4 × Magnetic field due to wire BC

Let CAB be the wire making an angle α, P be a point on the bisector of this angle situated at a separation x from the vertex A and d be the perpendicular distance of AC and AB from P.

From the diagram,

⇒

The induced magnetic field B due to wire AB

⇒

⇒

=

=

The magnetic field due to current in both the wires

=

=

Let us consider two angles made by points A and B with point O be θ₁ and θ₂ respectively.

θ₁ = θ₂ = θ

Let distance of the point from the wire

In triangle

Thus induced magnetic field due to AB is given by,

⇒

⇒

⇒

Similarly the induced magnetic field due to current in BC is given as:

Now, magnetic field due to CD = magnetic field due to current in wire AB =B

Also, magnetic field due to current in wire DA = magnetic field due to current in wire BC =B’

∴

a) Using the diagram,

For a polygon of n equal sides, the angle at the centre is .

⇒

Considering angle to be very small

Using Biot–Savart’s law for one side of the n-sided polygon,

⇒

(substituting the value of r)

(substituting the value of l)

For n-sided polygon

B’ =nB

⇒

b) When n → ∞, polygon tends to a circle with radius r and magnetic field will tend toward

By applying Kirchhoff’s Voltage Law, we can observe that the current in the circuit is zero

Now, since net current in the circuit = 0

Induced magnetic field is always proportional to the current flowing in the circuit. Hence,

Net magnetic field at the point P=0

Induced magnetic field at point P is independent of the values of the various resistances in the electric circuit.

Current on both wires i₁=i₂=20A

Force acting on 10 cm of the second wire, F =20 x 10^-5

∴ force per unit length =

Now, let the distance between the two wires be d. Thus, the force per unit length is given as:

⇒

⇒ distance d = 0.4m =40cm

-

Given: - Current flowing in each wire is i₁=i₂=i₃=10A

The magnetic force acting per unit length on wire due to the flow of the current is

So, for wire w_1,

= by wire w₂ + by wire w₃

= +

= 6.4 x 10^-4 N

And for wire w₂

= by wire w₁ - by wire w₃

= -

= 0

Similarly for wire w₃

= by wire w₁ + by wire w₂

= +

= 6.4 x 10^-4 N

Let x be the distance of the wire W₃ from the wire carrying a current of 10A.

Magnetic force acting per unit length on a wire due to a parallel current-carrying wire is given by

In this case as the wire W₃ experiences no magnetic force.

∴ by wire w₁ = by wire w₂

⇒

⇒ 10 – x = 4x

⇒ x = 2 cm

Thus, wire W₃ is placed 2 cm from the 10 A current wire.

Current will equally divide into three parts since wires AB, CD and EF have identical resistance, 10 A each.

So, the Magnetic force on wire AB due to wire CD is,

= = N/m

and Magnetic force on wire AB due to wire EF is,

= = N/m

Total force per unit length,

= + = =

Now, magnetic force on wire CD due to wire AB,

= = N/m

and magnetic force on wire CD due to wire EF is

= = N/m

= + = =

The current in the coil is 10A

There will be induction of magnetic force due to two current carrying wire placed in each other’s vicinity.

The force is given by

Where l is the length of the wire and i₁ i₂ are the currents flowing through the wires.

To balance the weight, the gravitational force should be equal to the magnetic force that is exerted.

Hence

Where λ is the linear mass density of the wire.

Hence, equating these forces, we get,

=

Substituting the numerical values, we get

Here, current in the wire is

Current in the loop,

Force on an element dx in the arm PQ a distance x away is

similarly, force on RS is,

Hence,.

B distance of PS from wire,

distance of PS from RQ,

Force on element PS is

towards right

Now distance of RQ from wire,

distance of RQ from PS,

similarly, force on RQ is

towards left

Net force,

Given:
No. of turns, n = 1
Magnitude of current, i = 5.00 A
the magnetic field at the center due to the current in the loop is given by

We know that

Now,

Where B=Magnetic Field= 0.2 mT

r= radius of loop

number of turns n =100 turns

Radius r = 5cm =0.05m

Magnetic field B =6.0 × 10^–3 T

Now,

i = 4.7740 A

no of revolutions = 3 × 10⁵

radius = 0.5 angstrom = 5×10⁻¹¹ m

Thus, the magnetic field at the centre due to the current in the loop is given by

We know that,

Here n=1,

I= =

and r= 5×10⁻¹¹ m

Hence,

The current flows into the circular coil and using the KCL, splits equally between the two arms because both the arms are equal in length and ae made up of the same material.

Hence the current in the upper section as well as the lower section will be i/2.

The current leaves the loop again and the resultant current leaving the coil will again be i.

The magnitude of the current is same and the direction is opposite at every point of time. Magnetic fields produced will hence also be zero as both the arms produce equivalent magnetic field which neutralize each other. Hence net magnetic field at the center will be 0

n_l = 100

n_s = 50

r_{l =} 10 cm

r_{s =} 5 cm

Both the coils have a common center and hence the effect due to the current flowing will depend on the direction of flow. We can calculate the individual results and then do the vector addition of both the forces to get the resultant magnetic fields

1) In the same sense

Field due to larger coil

Field due to smaller coil

As the direction is same and both the fields are acting in the same direction, that outside the plane of the paper, we can simply perform a scalar addition

Similarly for

2) Opposite sense

As the direction is opposite, we can simply perform scalar subtraction to get the resultant

∴

Given, for outer circle,

Number of turns ,

current,

Radius,

therefore, magnetic field is pointing west,

for inner circle,

Number of turns ,

current,

Radius,

therefore, magnetic field is pointing downwards,

.

Therefore, total magnetic field is

Given,

Radius,

Velocity of electron,

direction of motion,

Current in the loop,

Therefore, force,

Given:

Current carried by circular loop of radius R = I

Current carried by circular loop of radius r(<<R) = i

Diagram:

Formula used:

Magnetic field at the center of a circular loop=𝛍₀I/2r,

where I = current through the loop, r = radius of the circle,

𝛍₀ = 4π x 10^-7 kgm s^-2 A^-2

Magnetic torque acting on a loop of radius r = I (A X B), where I = current carried by the loop, A = area of loop, B = magnetic field

Therefore, magnetic field acting at the center of the loops due to the bigger loop of radius R(B) = μ₀I/2R.

Area of the smaller loop(A) = πr²

Hence, torque acting on the smaller loop = I (A X B), since current carried by smaller loop is i. Here A = area of smaller loop = πr² and B = magnetic field due to bigger loop.

Now, we know that A X B = |A||B|sinθ, where θ is the angle between the area vector of the smaller loop and the magnetic field of the larger loop. |A| = modulus value of area of smaller loop, |B| = modulus value of magnetic field due to larger loop.

Since the planes of the two loops are perpendicular to each other, θ = 90⁰.

Hence, torque acting on the smaller loop = i (|A||B|sin 90⁰) = i(|A||B|)

We know, |A| = πr², |B| =

Hence, torque on smaller loop = = (Answer)

Given:

Current carried by circular loop of radius r = i

Current carried by circular loop of radius R(>>r) = I

Angle between the planes of the two loops = 30⁰

Diagram:

Formula used:

Magnetic field at the center of a circular loop of radius r

M=,

where

I = current through the loop,

r = radius of the circle,

𝛍₀ = 4π x 10^-7 kgm s^-2 A^-2

Magnetic torque acting on a loop of radius r

where

I = current carried by the loop,

A = area of loop,

B = magnetic field,

F = force on periphery

Therefore, magnetic field at the center of the loops due to the bigger loop(B) =

Area of the smaller loop(A) = πr²

Therefore, torque on the smaller loop = i (A X B), since the smaller loop carries current i.

A = area of smaller loop, B = magnetic field due to bigger loop.

Now, we know that A X B = |A||B|sinθ,

Where

θ is the angle between the area vector of the smaller loop and the magnetic field of the larger loop.

|A| = magnitude of area of smaller loop,

|B| = magnitude of magnetic field due to larger loop.

Given:

The angle between the area vector of the smaller loop and the magnetic field of the larger loop is θ = 30⁰.

Hence, torque acting on the smaller loop = i(|A||B|sin 30⁰) =

We know, |A| = πr², |B| =

Hence,

Torque on smaller loop due to magnetic field,

Torque due to external force on its periphery,

Therefore

Where

F = external force,

r = perpendicular distance between the line of action of force and the point at which the torque is acting.

Here, F and r are perpendicular to each other, since the force is being applied to the periphery.

Here r = radius of smaller loop.

Hence,

which gives us

(Answer)

Given:

Radius of semicircular wire(r) = 10 cm = 0.1 m

Current carried by it(i) = 5 A

Formula used:

Magnetic field at the center of a semicircular loop =

Where

i = current carried by loop,

r = radius of loop,

𝛍₀ = magnetic permeability of vacuum = 4π x 10^-7 kg m s^-2 A^-2

Hence, required magnetic field(B) = Tesla(T) ,

where 𝛍₀ = 4π x 10^-7 kg m s^-2 A^-2

Given:

Radius of circular arc(r) = 10 cm = 0.1 m

Current carried by it(i) = 6 A

Angle subtended at the center = 1200

Formula used:

Magnetic field at the center of a circular arc,

Where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 kg m s^-2 A^-2,

i = current carried by the wire,

r = radius of the wire,

θ = angle subtended at the center of the arc

Diagram:

Hence, required magnetic field

where, i = 6 A, r = 0.1 m, θ = 120⁰ = 2π/3 radian, μ₀ = 4π x 10^-7 kg m s^-2 A^-2

Hence, B = = 1.256 x 10^-5 T = 12.56 μT (Ans)

For the magnetic field at the center to be zero, the magnetic field due to the circular loop has to be exactly equal and opposite to the long straight wire at the center.

Given:

Current carried by circular loop of radius r = i

Current carried by long straight wire = 4i

Formula used:

Magnetic field at the center of a circular loop due to current in the loop(B_l) =, which points into the plane of the paper.

Here

μ₀ = magnetic permeability of vacuum,

i = current carried by the loop,

r = radius of the loop

Magnetic field due to an infinitely long wire at a distance x from it(B_w) =, which points out of the plane of the paper.

Here

μ0 = magnetic permeability of vacuum,

i = current carried by the wire,

x = distance of the center of the circle from the wire

Hence, from given information,

, and

where B_l = magnetic field due to loop, B_w = magnetic field due to wire, r = radius of loop, x = distance of the center of the circle from the wire

Hence, for the magnetic field at the center to be 0,

=> =>

Hence, for the magnetic field to be 0, the wire has to be placed at a distance of 4r/π from the center of the circular loop. (Answer)

Given:

Number of turns(n) = 200

Radius of circular coil(r) = 10 cm = 0.1 m

Current carried by the coil(i) = 2 A

Formula used:

(a) Magnetic field at the center of a circular coil of n turns(B) =

M=,

where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1,

i = current carried by coil,

r = radius of coil

Hence, from the given information,

Magnetic field at the center of the given coil

= T = 2.51 x 10^-3 T = 2.51 mT

(b) The magnetic field at a distance point in the coil is given by B_p,

The magnetic field at the center of the coil is

Given: the magnetic field at the center is 1/2 of the initial

The magnetic field at the center of the coil is

On equation the magnetic field at the center and the magnetic field at the any distant point from the coil, we get

On equating the above equation, we get

(r² + x²)³ =4r⁶

We get

x² + r²= 4^1/3r²

x²=0.58 r²

x²=0.5× 100=58

x=± 7.66 cm

Magnetic field will drop to half of its value at the center if the distance of that point from the center of the coil along the axis of coil is equal to 7.66 cm.

Given:

Radius of circular loop(r) = 4 cm = 0.04 m

Current carried by it(i) = 5 A

Formula used:

Magnetic field at a distance x from the center of a current carrying circular loop(B),

Where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1,

i = current carried by coil,

r = radius of coil

(a) To find the magnetic field 3 cm above the loop:

Magnetic field,

Here, x = 3 cm = 0.03 m

Hence, substituting the given values:

= 4.019 x 10^-5 T (in the downward direction, by right hand thumb rule) (Ans)

(b) To find the magnetic field 3 cm below the loop:

Here, all the values are same as the previous part. Hence,

B = 4.019 x 10^-5 T (in the upward direction, by right hand thumb rule)

Given:

Charge of the ring(q) = 3.14 × 10^–6 C

Radius of circular ring(r) = 20 cm = 0.2 m

Angular velocity with which the ring rotates(w) = 60 rad s^-1

Diagram:

E₁ and E₂ denote the electric field due to two point elements 1 and 2 on the circumference of the circular loop. The resultant electric field due to them is E. Similarly, the resultant magnetic field due to the them is B.

Formula used:

Electric field due to a circular loop(E),

where q = charge carried by the ring, x = distance from the center, r = radius of the ring, ϵ₀ = electric permittivity of vacuum = 8.85 X 10^-12 N m² C^-2

Magnetic field due to a current carrying loop(B),

where μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1, i = current carried by the loop, r = radius of the loop, x = distance from the center

Now, i = q/t, where i = current, q = charge, t = time taken.

Angular velocity(w) = 60 rad/s (given)

Hence, time period of one revolution(T) = 2π/w = 2π/60 s

Hence, current(i) = charge per unit time = q/T = 60q/2π = 30q/π A, where q = charge, T = time period of revolution

Therefore, E/B = (2 x 3.14 × 10^–6 x 0.05 x 9 x 10⁹ x 3.14)/ (4 x 3.14 x 10^-7 x 30 x 3.14 × 10^-6 x 0.2²) = 1.88 x 10¹⁵ ms^-1 (Ans)

Formula used:

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

,

where B = magnetic field, dl = line element, μ₀ = magnetic permeability of vacuum, I = current enclosed.

Diagram:

(a) Now, since inside a conductor, the current enclosed is 0, in Ampere’s circuital law, we put I = 0.

Hence, the magnetic field at a distance r/2 from the surface inside the tube is 0. (Ans).

(b) Now, for any point outside the tube, the current enclosed will be given by Ampere’s circuital law:

Where

B = magnetic field,

dl = line element,

μ₀ = magnetic permeability of vacuum,

I = current enclosed.

Now, since we have to find the magnetic field at a distance r/2 from the surface,

we consider an Amperian loop of radius from the centre of the loop.

The total current enclosed will be i, since we are considering a loop outside the tube.

Hence,

,

since = circumference of the Amperian loop.

Hence, we get, B(at a distance r/2 from the surface outside the tube) =(Ans)

Given:

Inner radii of tube = a

Outer radii = b

Current distributed over its cross section = i

Diagram:

Formula used:

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

,

Where

B = magnetic field, dl = line element, μ₀ = magnetic permeability of vacuum, I = current enclosed.

(a) Inside a conducting tube, I (current enclosed) is 0.

Hence, by Ampere’s circuital law, = 0 => B = 0.

Hence, the magnetic field just inside the tube is 0.

(b) Just outside the tube, the distance from the center is b.

Hence, we consider an Amperian loop of radius b from the center.

Hence, from Ampere’s circuital law:

,

Where B = magnetic field,

dl = line element, μ₀ = magnetic permeability of vacuum, I = current enclosed.

In this case, ∮dl = 2πb (circumference of the loop of radius b)

Therefore, B x 2πb = μ_oi (since the total current enclosed is i).

Hence, the magnetic field at a point just outside the tube = μ_oi/2πb(Ans)

Given:

Radius of wire = b

Current distributed throughout the cross section = i

Diagram:

Formula used:

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

, where B = magnetic field, dl = line element, μ₀ = magnetic permeability of vacuum, I = current enclosed.

Since the current i is uniformly distributed throughout the cross section of the wire,

For an amperian loop of radius a(a<b) from the center of the wire,

I (current enclosed)

Where

i= total current distributed throughout the wire,

b = radius of the wire,

a = distance from the axis at which the magnetic field is to be found.

Hence, from Ampere’s circuital law,

since 2πa = circumference of Amperian loop of radius a, and I = enclosed current = ia²/b².

Therefore, magnetic field at a distance a from the axis inside the wire,

Given:

Radius of wire(r) = 10 cm = 0.1 m

Current carried by it(i) = 5 A

Diagram:

Formula used:

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

,

Where B = magnetic field, dl = line element, μ₀ = magnetic permeability of vacuum, I = current enclosed.

(a) Since the current is uniformly distributed over the cross section of the wire, at a distance of 2 cm (0.02 m) from the axis (inside the wire), I (current enclosed)

=

Where,

A = 0.2 A, where 0.1 m is the radius of the wire.

Hence, from Ampere’s circuital law,

where μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1, B = magnetic field, dl = line element, I = current enclosed

Substituting the values, we get

B x 2π x 0.02 = 4π x 10^-7 x 0.2 (since 2π x 0.02 = circumference of the loop of radius 2 cm)

=> Magnetic field at a distance of 2 cm from the axis

= 2 x 10^-6 T = 2μT (Ans)

(b) At a distance of 10 cm from the axis, we are basically on the surface of the wire. Hence the whole current of 5 A is enclosed by it.

Hence, from Ampere’s circuital ,,

where μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1, B = magnetic field, dl = line element, I = current enclosed

Here, I = 5 A.

Substituting the values, we get

B X 2π x 0.1 = 4π x 10^-7 x 5(since 2π x 0.1 is the circumference of the loop of radius 10 cm)

=> Magnetic field at a distance of 10 cm from the axis

= 10^-5 T = 10μT (Ans)

(c) At a distance of 20 cm from the axis, we are outside the wire. Hence the whole current of 5 A is enclosed by it.

Hence, from Ampere’s circuital law, ,

where μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1, B = magnetic field, dl = line element, I = current enclosed

Here, I = 5 A.

Substituting the values, we get

B X 2π x 0.2 = 4π x 10^-7 x 5(since 2π x 0.2 is the circumference of the loop of radius 20 cm)

=> Magnetic field at a distance of 20 cm from the axis

= 5 x 10^-6 T = 5μT (Ans)

Graph for B vs x for 0 < x < 20 cm:

Half of the given loop is in a region with zero magnetic field, while half of it is in a region with magnetic field B.

Now, we take the Amperian loop PQRS.

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

where

B = magnetic field,

dl = line element,

μ₀ = magnetic permeability of vacuum,

I = current enclosed.

Hence, I (current enclosed) is not 0, but is finite (since half of it lies in a region of magnetic field B). Hence, this states the magnetic field inside the loop PQRS is not zero.

However, it is obvious that the magnetic field at any point in the region outside the magnetic field lines = 0.

These are two contradictory statements.

Hence, such a field is not possible. (proved)

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

, where B = magnetic field, dl = line element, μ₀ = magnetic permeability of vacuum, I = current enclosed.

(i) At point P, the current enclosed(I) = 0.

Hence, from Ampere’s circuital law, ∮B.dl = 0 => B (magnetic field at P) = 0. (Ans)

(ii) Point Q is in between the two strips carrying current.

Hence, from Ampere’s circuital law, we consider only B.dl for the strip of width dl carrying surface current K. Hence, I (current enclosed by the strip) = Kdl.

Therefore, B.dl = μ₀Kdl => B (magnetic field at Q) = μ₀K (Ans)

(iii) At point Q, again, current enclosed(I) = 0.

Hence, from Ampere’s circuital law, ∮B.dl = 0 => B (magnetic field at Q) = 0. (Ans)

Given:

Charge of the particle = q

Mass of the particle = m

Radius of circle described by the particle between the two plates = r

Formula used:

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

where B = magnetic field, dl = line element, μ₀ = magnetic permeability of vacuum, I = current enclosed.

Now, from the previous problem, we found out that the magnetic field(B) at point Q is μ₀K, where μ₀ = magnetic permeability of vacuum, K = surface current.

Also, we know that the velocity of a particle in a magnetic field is given as

Where

q = charge of particle, B = magnetic field, r = radius of circle described by particle, m = mass of particle.

Substituting the value of B, we get

Velocity of the particle is

Given:

Current(i) = 5 A

Magnetic field(B) = 3.14 × 10^–2 T

Formula used:

Magnetic field inside a long solenoid B = μ₀ni,

Where

μ0 = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1, n = number of turns per unit length, i = current carried by the wire

Hence, from the given information:

3.14 × 10^–2 = 4π x 10^-7 x n x 5

=> n = 5000 turns/m

Therefore, number of turns per unit length of the solenoid = 5000(Ans)

Given:

Radius of cylinder(r) = 0.5 mm = 5 x 10^-4 m

Current(i) = 5 A

Formula used:

Magnetic field inside a long solenoid B = μ₀ni,

where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1, n = number of turns per unit length, i = current carried by the wire

Now, width of each turn = diameter of the wire = (0.5 x 2) mm = 1 mm = 10^-3 m.

Hence, number of turns per unit length(n) (per m) =10³ = 1000

Therefore, substituting the values:

B (magnetic field at the center) = 4π x 10^-7 x 1000 x 5 T

= 6.28 x 10^-3 T (Ans)

Given:

Resistance per unit length(R₀) = 0.01 ohm

Total number of turns(N) = 400

Radius of the wire(r) = 1 cm = 0.01 m

Length of the wire(l) = 20 cm = 0.2 m

Magnetic field near the center(B) = 1.0×10^–2 T

Formula used:

Magnetic field at the center of a solenoid, B = μ0ni,

Where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1,

n = number of turns per unit length,

i = current carried by the wire

Emf(E) = iR,

where i = current, R = resistance

Now, n(number of turns per unit length) = Total number of turns(N)/Length of wire(l) = = 2000

Total resistance (R of all turns) = R₀ x 2πr x 400 = (0.01 x 2π x 0.01 x 400) = 0.25 ohm

Now, substituting the given values:

1.0×10^-2 = 4π x 10^-7 x 2000 x i => i = 3.98 A

Therefore, Emf(E) = iR = (3.98 X 0.25) = 0.995 V which is almost equal to 1 V.

Emf of the battery = 1 V (Ans)

Given

Radius of solenoid is a

Length of the solenoid is l

Number of turns per unit length is n

The current in the circular loop is indx

The magnetic field due to the circular ring at the any distance from the ring is

Where

μ_o is the permeability of the free space,

I is the current in the ring

r is the radius of the loop

x is the distance from the loop at which the magnetic field to be found.

Integrating the above equation for x-0 and r

On solving the above equation, we get

The magnetic field at the center of the solenoid due to this circular current

(b) When l≫a

Putting in the equation (i), we get

Putting the values in the above formula, we get

B=μ₀ n I

When a ≫ l

Putting in the equation (i), we get

Putting the above value in the equation (i), we get

Hence, proved if l >> a, the field tends to B = μ₀ni and

if a >> l, the field tends to B, .

Given:

Current(i) = 2 A

Frequency(f) = 1.00 × 10⁸ rev s^–1.

Formula used:

Magnetic field inside a solenoid(B) = μ₀ni,

Where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1,

n = number of turns per unit length,

i = current carried by the wire

Now, charge of electron(q) = 1.6 x 10^-19 C

Mass of electron(m) = 9.1 x 10^-31 kg

Now, frequency of the particle in uniform circular motion in a magnetic field is given by

f = ,

Where

q is the charge,

B is the magnetic field,

m is the mass of particle

Hence, B =

Substituting the given values, we obtain B as,

Hence, B = μ₀ni,

Where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1,

n = number of turns per unit length,

i = current carried by the wire

=> Number of turns per meter(n) = = 0.0036/(4π x 10^-7 x 2) = 1420 turns/meter. (Ans)

Given:

Number of turns per unit length = n

Current = i

Radius = r

Charge of particle = q

Mass = m

Formula used:

Magnetic field inside a solenoid(B) = μ₀ni,

where

μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1,

n = number of turns per unit length,

i = current carried by the wire

Now, when a particle is projected perpendicular to a magnetic field, it describes a circle. Now, for the particle to not strike to solenoid, the required radius is r/2.

Since it is moving in a circular path, centripetal acceleration = mv²/r,

Where

m = mass of particle,

v = velocity,

r = radius

Force due to the magnetic field = qvB,

Where

q = charge of particle,

v = velocity,

B = magnetic field.

Here, r => r/2.

The centripetal and magnetic forces balance each other.

Hence, using the given data:

=> v = qμ₀nir/2m (Ans)

Given:

Number of turns per unit length = n

Surface current of sheet = K

Formula used:

Ampere’s circuital law states that the line integral of the magnetic field for a closed surface is μ₀ times the current enclosed by the surface.

,

Where

B = magnetic field,

dl = line element,

μ₀ = magnetic permeability of vacuum,

I = current in the circuit.

(a) Since the magnetic field near the center of the solenoid is 0, we infer that the magnetic field due to the solenoid = magnetic field due to the sheet

For the sheet, using Ampere’s circuital law, we get:

B x 2l = μ₀ x Kl, where Kl = current enclosed by sheet and we take 2l because of the two surfaces of the sheet.

=> B_solenoid = μ₀K/2

Now, magnetic field due to a solenoid B_solenoid = μ₀ni, where μ₀ = magnetic permeability of vacuum, i = current, n = number of turns per unit length

Hence, μ₀K/2 = μ₀ni => i (current in solenoid) = K/2n (Ans)

(b) Now, if the axes of the solenoid and the plate are perpendicular to each other, their magnetic fields are also perpendicular to each other.

Hence, net magnetic field at center,

Given:

Capacitance(C) = 100 μF = 10^-4 F

Initial Voltage of battery(V) = 20 V

Number of turns per meter of solenoid(n) = 4000

Final potential difference(V’) = 90% of 20 V = 18 V

Time taken(t) = 2 s

Formula used:

Initial charge stored in capacitor(Q) = CV,

where C = capacitance, V = potential difference

Therefore Q = (10^-4 x 20) C = 2 x 10^-3 C

New potential difference = V’ = 18 V

Therefore, New charge Q’ = CV’ = (10^-4 x 18) = 1.8 x 10^-3 C

Hence, current flowing in conductor(i) = (Q-Q’)/t, where Q = initial charge, Q’ = final charge, t = time taken

Therefore, i = = 10^-4 A

Thus, average magnetic field at the center of solenoid = B = μ₀ni,

where μ₀ = magnetic permeability of vacuum = 4π x 10^-7 T m A^-1, n = number of turns per unit length, i = current carried by the wire

Hence, from given data, B = (4π x 10^-7 x 4000 x 10^-4) T

= 0.5 x 10-6 T (Ans)