ELECTROMAGNETIC INDUCTION AND ALTERNATING CURRENTS

6 )

A solenoid of 1.5 m long with inner diameter of 4 cm has three layers of windings of 1000 turns each & carries a current of 2 A. The magnetic flux through a cross section of the solenoid is nearly.....

2.5 × 10^–7 wb

6.31 × 10^–6 wb

2.1 × 10^–6 wb

4.1 × 10^–5 wb

View Solution

Answer : Free Physics Questions Online Mock Test Series By Cracktest

L = 1.5 m

I = 2A

Total number of turns = 3 × 1000 = 3000

∴ Turns per unit meter = n = [(3000) / (1.5)] = 2000

r = [(4 cm) / 2] = 2 × 10^–2 m

As ɸ = AB cos θ = AB cos 0

ɸ = AB

put B = μ_onI & A = πr² we get

ɸ = π r² × μ_onI = π × (2 × 10^–2)² × 4π × 10^–7 × 2000 × 2

∴ θ = 6.31 × 10^–6 wb

7 )

A coil has an area of 0.05 m² & it has 800 turns. It is placed perpendicularly in a magnetic field of strength, 4 × 10^–5 Wb/m^–2. It is rotated through 90° in 0.1 sec. The average emf induced in the coil is...

0.056 v

0.046 v

0.026 v

0.016 v

View Solution

Answer :

A = 0.05 m²

N = 800

B = 4 × 10^–5 Wb/m²

t = 0.1 sec

e = [{NAB} / {t}]

∴ e = [{800 × 0.05 × 4 × 10^–5} / {0.1}]

e = 0.016 v

8 )

A bar magnet is moving along the common axis of two coils A &B towards A. current is induced in

Only A

Only B

both A& B in same direction

both A & B in opposite direction

View Solution

Answer :

When bar is moving along common axis of two coils then current is

included in both coils. Because when magnet is moving, there will be

change of magnetic flux linked with the circuit. And change of flux will

result into emf and finally current is induced in both the coils and direction

in one coil will be clockwise and in other coil it will be anticlockwise.

9 )

A coil having n turns & resistance RΩ is connected with a galvanometer of resistance 4RΩ. This combination is moved from a magnetic field W₁ Wb to W₂ Wb in t second. The induced current in the circuit is....

Select correct option from the following options

(a) – [{W₂ – W₁} / {5Rnt}] (b) – n [{W₂ – W₁} / {5Rt}]

View Solution

Answer :

Turn = n

Resistance = RΩ

galvanometer resistance = 4RΩ.

e = – N (dɸ / dt) = – n [{W₂ – W₁} / {t}]

induced current = [e / {Total resistance}]

∴ I = [e / {R + 4R}] = (e / 5R)

∴ I = – n [{W₂ – W₁} / {5Rt}]

10 )

When a train travels with a speed 360 km h^–1 along the track separated by 1 meter. The vertical component of earth's magnetic field is 0.1 × 10^–4 T .What is the value of induced emf between the rails?

10^–2 V

10^–3 V

10^–4 V

1 V

View Solution

Answer :

Velocity of train = 360 kmph = [{360 × 100°} / {360°}] m/s = 100 m/s

The train cuts vertical component of earth’s magnetic field.

∴ emf is induced across the axie is e = BVℓ

here

ℓ = 1 m,

B = 0.1 × 10^–4 T

∴ e = 0.1 × 10^–4 × 100 × 1

e = 10^–3 V

Topic Categories