Electromagnetic Induction

The direction of induced current will be given by Lenz’s law.

According to the Lenz’s law: The direction of induced current by change in magnetic field (Faraday’s induction) will be such that it opposes the change that caused it.
As seen from the given figures, the direction of the induced current is shown when the north pole of the magnet is moved towards the closed loop or away from the closed loop.

According to the Lenz’s law: The direction of induced current by change in magnetic field (Faraday’s induction) will be such that it opposes the change that caused it.

(a) In the figure given below, the direction of magnetic field is perpendicular into the paper.

Now, when we change the shape of irregular loop into a circular loop, then the area of the loop will increase. Because, area of circle is considered to be greatest.

Since, area of loop increases, ∴ the magnetic flux will also increase.

Now, the current will be induced in the loop such that the magnetic flux decreases. And, magnetic field will decrease when the area will decrease i.e. the wires should be pulled in the inwards directions, to decrease the net magnetic field.

∴ the current will be along “adcba”.

(b) In the figure given below, the direction of magnetic field is perpendicular coming out of the paper.

Now, when circular loop is transformed into a narrow straight wire, then the area of the loop will decrease. Since, area of loop decreases, ∴ the magnetic flux will also decrease.

Now, the current will be induced in the loop such that the net magnetic flux increases in the wire. ∴ the current flow will be along “adcba” which produces the induced magnetic field outwards the paper.

Given:

No. of turns in the solenoid coil, N = 15 turns/cm = 15000 turns/m

(∵ 1m = 100 cm)

∴ no. of turns in the solenoid coil per unit length, n = 15000 turns

Area of the solenoid coil = 2.0 cm²

In m², Area will be 2 × 10 ^{- 4} m²

Since current carried by the solenoid coil changes from 2.0 to 4.0 A in 0.1 seconds

Therefore, change in the current in the coil of solenoid = final current – initial current

di = 4.0 – 2.0 = 2.0 A

The change in time “dt” is given as ,dt = 0.1 seconds

Applying Faraday’s law, induced e.m.f can be calculated as follows:

…(1)

Where Ф is flux induced in the loop

And, flux is given as, Ф = BA

Where “B” is the magnetic field and “A” is the area of the loop.

For a solenoid, B= μ₀nI

Therefore, the equation (1) becomes:

Or

Or E = Aμ₀ n (dI/dt)…(2)

∵ A, μ_o and n are constants

Substituting the values in equation (2), we get,

⇒ E = 7.54 × 10 ^{- 6} V

Or E = 7.54 μV

(∵ 10^-6 = μ)

Hence, the induced voltage in the loop is 7.54 μV

Given:

Length of rectangular wire = 8 cm

Breadth of rectangular wire = 2 cm

The area of the rectangular wire can be calculated as follows:

A = l × b

A = 8 cm × 2 cm = 16 cm²

In m², area will be A = 16 × 10 ^{- 4} m²

Magnitude of magnetic field B = 0.3T

Velocity of the loop, v =1 cms ^{- 1} = 0.01ms ^{- 1}

(a) The emf developed across the cut can be calculated as follows:

e = Blv …(1)

where B is the magnetic field

l is the length of the loop and,

v is the velocity of the rectangular loop

substituting the values in equation (1), we get,

e = (0.3 T × 0.08m × 0.01ms ^{- 1})

e = 2.4 × 10 ^{- 4} V

Time taken for travelling along the width of the loop can be calculated as follows:

T =distance travelled/velocity with which distance is travelled

T = b(width of the loop)/v (velocity)

T = 0.02 m/0.01 ms ^{- 1}

T= 2 seconds

This means that the induced voltage will last for very short direction of 2 seconds.

(b) The emf developed across the cut if the velocity of the loop is 1 cms^–1 in a direction normal to the shortest side of the loop can be calculated as follows:

e = Bbv…(2)

where B is the magnetic field

b is the width of the loop and,

v is the velocity of the rectangular loop

substituting the values in equation (2), we get,

e = (0.3 T × 0.02 m × 0.01 ms ^{- 1})

e = 0.6 × 10 ^{- 4} V

Time taken for travelling along the width of the loop can be calculated as follows:

T =distance travelled/velocity with which distance is travelled

T = l(length of the loop)/v

T = 0.08 m/0.01ms ^{- 1}

T = 8 seconds

This means that the induced voltage will last for 8 seconds.

Given:

Length of the metallic rod = 1.0 m

Angular frequency = 400 rads ^{- 1}

Strength of magnetic field B= 0.5 T

Since it is clear that one end of the metallic rod has zero linear velocity whereas the other end of the rod has linear velocity of lω

∴ The average linear velocity of the metallic rod can be calculated as follows:

v = (lω + 0)/2

⇒ v = lω /2

The e.m.f. developed between the centre and the ring can be calculated as follows:

e = Blv ...(1)

where B is the magnetic field

l is the length of the loop and,

v is the velocity of the rectangular loop

⇒ Substituting the value of v in above equation, we get

e = B × l × (lω/2)

e = (Bl²ω)/2

e = (0.5 T × 1m² × 400 rads ^{- 1})/2

On calculating, we get

e = 100 V

Hence, the emf developed across the ring is 100 V.

Given:

Radius of the circular coil r = 8.0 cm = 0.08 m

No. of turns in the coil, N =20

Angular frequency ω = 50 rad s ^{- 1}

Magnitude of magnetic field B =3.0 × 10^–2 T

Resistance of the closed loop = 10 Ω

The area of the coil can be calculated using the formula

A =πr²

A = 3.14 × (0.08 m)²

The maximum induced e.m.f is calculated as follows:

e = N ω A B

e = 20 × 50 rad s ^{- 1} × 3.14 × (0.08 m)² × 3.0 × 10^–2 T

On calculating we get

e = 0.603 V

Since for a full cycle, the average induced emf will be zero.

The maximum current for the circular coil can be calculated as follows:

I = e/R

Substituting the values

I = 0.603 V/10 Ω

I = 0.0603A

The average power loss due to Joule’s heating effect in the circular coil will be given by:

P = (eI)/2

P = (0.603 V × 0.0603 A)/2

On calculating we get

P = 0.018 W

A torque was produced due to current induced in the coil. This torque opposes the rotation of the circular coil. Since a rotor is the external agent in the coil and it should apply the torque so that the coil keeps rotating in the uniform manner. Therefore, the dissipated power comes from the external rotor.

Given:

Length of the wire, l = 10 m

Speed of wire, v = 5.0 ms ^{- 1}

Magnetic field B = 0.30 × 10^–4 Wb m^–2

The diagram is:

(a) The e.m.f induced in the wire can be calculated as follows:

e = Blv ...(1)

where B is the magnetic field

l is the length of the loop and,

v is the velocity of the rectangular loop

substituting the values in above equation, we get,

e = 0.30 × 10^–4 Wb-m ^{- 2} × 10 m × 5ms^-1

e = 1.5 × 10^–3 V

(b) The direction of the induced emf will be from west to east in accordance with the Fleming’s right - hand rule.

According to Fleming’s right - hand rule: When the right hand is held with the thumb, first finger and second finger mutually perpendicular to each other, as shown below.

Then, the thumb points in the direction in which the conductor moves; the first finger gives the direction of the magnetic field (N ⇒ S); and the second finger gives the direction of the induced current.

(c) The east end of the straight wire will be having high electrical potential.

Given: Initial current I₁ = 5.0 A

Final current I₂ = 0.0 A

Time = 0.1 seconds

Average e.m.f = 200 V

Change in the current can be calculated as:

dI = I₁ - I₂ = 5.0A – 0.0A

dI = 5 A

The self-conductance of the circuit is calculated using the formula:

or

substituting the values, we get

∴ L = 4 Henry

Given: Mutual conductance, μ = 1.5 H

Initial current I₁= 0

Final current I₂ = 20 A

Time taken t = 0.5 seconds

Change in the current can be calculated as:

dI = I₁ - I₂ = 20 – 0.0

dI = 20 A

The mutual conductance of the circuit is calculated using the formula:

or

As we know that

Substituting it in equation, we get

Given: Speed of the jet plane towards west = 1800 km/h

In m/s, speed of the jet plane towards west = 500 m/s

Length of the span of the jet plane = 25 m

Magnetic field = 5 × 10^–4 T

Dip angle =30°

The vertical component of Earth’s magnetic field can be calculated as follows:

B_v= B sinδ

On substituting the values, we get,

B_v= 5 × 10^–4 T × sin30°

⇒ B_v= 5 × 10^–4 T × 1/2

∴ B_v = 2.5 × 10^–4 Tesla

The voltage difference developed between the ends of the jet can be calculated using the formula:

e = B_v × l × v

On substituting the values, we get

e =2.5×10^–4 T× 25m × 500m/sec

on calculating, we get

e = 3.125 V

∴ the voltage difference developed between the ends of the wings is approximately 3V.

Given: Length of rectangular wire = 8 cm

Breadth of rectangular wire = 2 cm

The area of the rectangular wire can be calculated as follows:

A = l × b

A = 8cm × 2cm = 16 cm²= 16 × 10^-4 m²

Initial Magnitude of magnetic field B = 0.3T

Velocity of the loop, v =1 cms^-1

The rate of decrease of the magnetic field i.e.

The e.m.f developed in the rectangular loop is given as:

dФ is the change in the flux across the loop = AB

⇒

Or

Therefore, substituting the values in above equation, we get:

e = 16×10^-4 m²× 0.02 Ts^-1

⇒ e = 0.32 × 10^-4V

The current induced in the loop can be calculated as follows:

i = e/r

i = (0.32 × 10 ^{- 4})/1.6

⇒ i = 2 × 10 ^{- 5}A

The Power loss or dissipation can be calculated as follows:

P = i²R

Where, R is resistance = 1.6Ω

⇒ P = (2 × 10^-5A)² × 1.6Ω

⇒ P = 6.4×10^-10 W

The source of the power is an external agent. This is due to change in the magnetic field with time.

Given: side of square loop = 12 cm

In metre, the side of square loop = 0.12m

Area of the square loop = 0.12 × 0.12 = 144 × 10 ^{- 4}m²

Velocity of the loop = 8 cms^-1

In metre/s, the velocity of the loop, v= 0.08m/s

Since gradient of electric field is along negative x - direction

In Tm ^{- 1},

∴ The rate of decrease of magnetic field in time is given as:

Resistance of the square loop = 4.50 mΩ =4.5 × 10 ^{- 3}Ω

Due to change in the motion of the square loop in presence of non - uniform magnetic field, the decrease in the magnetic flux is given as:

Due to explicit time variation in magnetic field, the rate of change of flux is given by:

Substituting values in above, we get

The total emf induced in the square loop is :

e =1.44 × 10^–5 V+ 11.52× 10^-5 V

e = 12.96 × 10^-5 V

The induced current can be written as:

i =e/R

⇒

⇒ i = 2.88 × 10 ^{- 2}A

The current direction will be in the way that there will be increase in the flux along positive z direction.

Given: Area of small flat search coil = 2 cm²

Total number of turns in the coil n = 25

Total charge measured by the ballistic galvanometer = 7.5 mC=7.5 ×10^-3C

combined resistance of the coil and the galvanometer R= 0.50

_{The current induced in the coil can be calculated as follows:}

_{i = e/R}

_also,

substituting in the equation we get

⇒ …………..(i)

Initial flux through the coil, Ф₁ = AB

Final flux through the coil, Ф₂= 0

Integrating equation (i) on both sides, we get

Total charge

Therefore,

⇒ Q = (NBA)/R

Therefore, B = (QR)/NA

So, substituting all the values in above equation, We get

On solving, we get

B = 0.75 T

Given: Length of the rod = 15 cm

Length of rod in metres l = 0.15 m

Magnetic field strength B = 0.50 T

Resistance of the loop R = 9 mΩ = 9 × 10 ^{- 3} Ω

Emf induced in the loop = 9 mV

Speed of the rod, v= 12 cm/s

Speed of the rod in m/s, v= 0.12m/s

It is shown as below:

(a) Induced emf can be calculated as follows:

e = Bvl

e = 0.50T× 0.12ms^-1× 0.15m

e = 9× 10^-3V = 9mV

The polarity of the induced emf is such that the end P is positive and end Q is negative.

(b) Yes, there is an excess charge built up at the ends of the rods when K is open. When K is closed, the excess charge present will be maintained by the flow of current continuously.

(c) The net force on the rod PQ is zero still they experience magnetic force due to motion of the rod magnetic force is cancelled by the electric force set up. This happens because of the excess charge on both the opposite ends of the rod.

(d) The retarding force acting on the rod when K is closed can be calculated as follows:

F = iBl……………(i)

Where i= induced emf(e)/Resistance (RΩ)

I = 9mV/9×10^-3Ω

I= 1A

Substituting the values in equation (i), we get

F = 1A× 0.50T× 0.15m

F = 75×10^-3 N

(e) Power required when K is closed:

Power = Fv

P = 75×10^-3 N× 0.12ms^-1

P = 9×10 ^{- 3} W

Power = 9 mW

When key K is open, there is no power required.

(f) Power dissipated H = I²R

H = 1A ²× 9×10^-3Ω = 9 mW

9mW power is dissipated as heat in the closed circuit and the source of this power is external agent.

(g) The induced emf in the moving rod if the magnetic field is parallel to the rails instead of being perpendicular is zero because the motion of the rod doesn’t cross across the lines and therefore induced emf will be zero.

Given: Length of the solenoid = 30 cm=0.30 m

Area of cross - section = 25 cm² = 25m× 10^-4m^-2

No. of turns in the solenoid = 500

Current, i = 2.5A

Time for which the current flows t = 10^–3 s

The average back emf induced across the ends of the open switch in the circuit is as follows:

………………(i)

where dФ is the change in the flux and is equal to NAB

B is the magnetic field = (μ₀Ni)/l

Substituting in equation (i), we get

Substituting the values in above equation , we get

On solving, we get

e = 6.5 V

∴ the emf induced in the solenoid is 6.5V

Let a small element dy in the loop at a distance of y as shown in the given figure:

Magnetic field associated with the the element taken i.e. dy can be calculated as follows:

dФ = BdA ………………..(i)

where dA is the area of the element dy i.e. ady

B is the magnetic field intensity of the element dy at a distance of y and can be written as:

Putting in the equation (i)

Putting limits of y from x to a + x

Mutual conductance is given by the relation:

Ф = MI

Where MI =

Or M =

The emf induced in the coil is given as:

e = B’ av

e =

substituting the given values in above relation, we get

e =

On calculating, we get

e = 5 × 10^-5V

Line charge per unit length is given by the following relation:

Total charge/ length = Q/2π r

Where r is the distance of the point within the wheel

Let m and R are mass and radius of the wheel

Magnetic field is given by the following relation:

The magnetic force at balanced by the centripetal force at a distance of r

i.e. BQv = mv²/r

v is the linear velocity of the wheel and is equal to v= 2π rλ

⇒ B × 2π rλ = mv/r

Or

As we know that angular velocity is given as: ω = v/r

⇒ w =

When R> a and r ≤ a

ω =