Electromagnetic Induction Test

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Electromagnetic Induction Test - 34

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Weekly Quiz Competition

Question 1
1 / -0

A small circular coil of radius 1 cm and number of turns 100 is placed inside a long solenoid of radius 5 cm and number of turns 8 per cm. The axis of the coil is parallel to the solenoid axis. Then, the coefficient of mutual inductance of the two coils is (in milli Henry)

A
0.032

B
0.064

C
0.016

D
zero

Solution

Field inside solenoid = $$B= \mu N i $$

N - number of turns per length in solenoid $$= 800/m$$

Flux linking the smaller coil = $$\lambda= 100 \times \pi r^2 B = 100 \times \pi r^2 \mu N i $$

mutual inductance = $$ M=\dfrac{\lambda}{i} = 100 \times \pi r^2 \mu N = 3.15 m H$$
Question 2
1 / -0

An athlete is running at a speed of 30 kmh$$^{-1}$$ towards east, holding a 3 m metallic rod horizontally. The horizontal component of the earth, magnetic field in this region is $$3 \times 10^{-4} $$ Tesla and the angle of dip is 30$$^o$$. Then, the emf induced across the ends of the rod is

A
7.5 mV

B
4.3 mV

C
zero

D
13 mV

Solution

Component of field in the vertical direction = $$ B_|=3 \times 10^{-4} tan( 30)$$
vertical component of field is perpendicular to both velocity and length of rod
emf = $$ B_| l v = 4.3 mV$$
Question 3
1 / -0

An ac generator consists of a coil of 200 turns, 100 cm in diameter. If the coil rotates at 500 rpm in a magnetic field of 0.25T, then the maximum induced emf

A
20.6 V

B
51.7 V

C
4.1 V

D
None of the above

Solution

$$emf =$$ rate of change of flux

$$= N \times area \times B \times \omega$$

$$= 200 \times \pi 0.5^2 \times 0.25 \times 500\times 2\pi / 60 $$

$$ = 2056.17 V $$
Question 4
1 / -0

In an a.c. generator the speed at which the coil rotates is doubled. How would this affect the frequency of output voltage ?

A
frequency is doubled

B
frequency is halved

C
frequency remains same

D
cant say

Solution

AC generators are made up of one or more magnetic pole pairs corresponding to their purposes.
The frequency of an AC generator is dependent on the number of pairs of poles and the speed of the coil rotation. The frequency of the induced AC current depends on the revolution of the coil since, rate of revolution is Revolutions Per Minute (RPM) and frequency is expressed as cycles per second, the frequency of the induced AC current will be the rate of revolution of the coil divided by 60 seconds.
That is, $$F=RPM\times number \ of\ pole\ pairs\div 60$$
When the Revolutions per minute (RPM) is doubled, then, obviously the frequency of the ac current will also be doubled as the RPM is multiplied with the number of pole pairs.
Question 5
1 / -0

Complete the following sentence:
The current is induced in a closed circuit only if there is _________.

A
change in number of magnetic field lines linked with the circuit.

B
no change in number of magnetic field lines linked with the circuit.

C
change in number of gravitational field lines linked with the circuit.

D
no change in number of gravitational field lines linked with the circuit.

Solution

The current is induced in a closed circuit only if there is a change in number of magnetic field lines in the circuit due to electromagnetic induction according to Faraday's law.
Question 6
1 / -0

A car moves up a plane road. The induced emf in the axle connecting the two wheels is maximum when it moves

A
At the poles

B
At equator

C
Remains stationary

D
No emf is induced at all

Solution

Induced emf in the axle $$ =B l v$$
$$v -$$ velocity of car
$$l -$$ length of car
$$B -$$ component of magnetic field perpendicular to both l and v .
That is B is the vertical component of magnetic field.
Vertical component of magnetic field is maximum at the poles.
Therefore emf induced in the axle will be maximum at the poles.
Question 7
1 / -0

An airplane with wingspan $$50 m$$ is flying horizontally with a speed of $$360 km hr^{-1}$$ over a place where the vertical component of the earth's magnetic field is $$2 \times 10^{-2}$$. The potential difference between the tips of the wings would be

A
$$100 V$$

B
$$1.0 V$$

C
$$0.2 V$$

D
$$0.01 V$$

Solution

emf between the wing tips $$= B l v\\$$

$$ = 0.02 \times 50 \times 360 \times \dfrac{1000}{3600}\\$$

$$ = 100 V $$
Question 8
1 / -0

One conducting U tube can slide inside another as shown in figure, maintaining electric contacts between the tubes. The magnetic field B is perpendicular to the plane of the figure. If each tube moves towards the other at a constant speed v, then the emf induced in the circuit in terms of B, $$l$$ and v where $$l$$ is the width of each tube, will be

A
-B$$l$$v

B
B$$l$$v

C
2 B$$l$$v

D
zero

Solution

Relative velocity $$ = v + v = 2v$$
Therefore, $$emf = Bl (2v)$$
Question 9
1 / -0

What determines the frequency of a.c. produced in a generator?

A
The number of rotations of the coil in one second.

B
the speed of rotation of the coil

C
both A & B

D
None of the above

Solution

Frequency of a.c, $$\nu = \dfrac{w}{2\pi}$$, where $$w$$ is the speed of rotation
Also, frequency is defined as the number of rotations that a coil rotates in one second.
Question 10
1 / -0

Two circular coils can be arranged in any of the three situations shown in the figure. Their mutual inductance will be

A
maximum in situation (A)

B
maximum in situation (B)

C
maximum in situation (C)

D
the same in all situations

Solution

The mutual inductance $$M$$ is given by, $$\phi=MI$$, where $$\phi$$ is the flux through a coil due to the current $$i$$ in another coil. The flux through an area $$\vec{A}$$. The $$\vec{B}$$ and $$\vec{A}$$ are parallel in configuration (A) but perpendicular in configuration (B) and configuration (C). Hence, the flux and mutual inductance are maximum in configuration (A).

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Electromagnetic Induction Test - 34

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Electromagnetic Induction Test - 34

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Question 1

0.032

0.064

0.016

zero

Solution

Question 2

7.5 mV

4.3 mV

zero

13 mV

Solution

Question 3

20.6 V

51.7 V

4.1 V

None of the above

Solution

Question 4

frequency is doubled

frequency is halved

frequency remains same

cant say

Solution

Question 5

change in number of magnetic field lines linked with the circuit.

no change in number of magnetic field lines linked with the circuit.

change in number of gravitational field lines linked with the circuit.

no change in number of gravitational field lines linked with the circuit.

Solution

Question 6

At the poles

At equator

Remains stationary

No emf is induced at all

Solution

Question 7

$$100 V$$

$$1.0 V$$

$$0.2 V$$

$$0.01 V$$

Solution

Question 8

-B$$l$$v

B$$l$$v

2 B$$l$$v

zero

Solution

Question 9

The number of rotations of the coil in one second.

the speed of rotation of the coil

both A & B

None of the above

Solution

Question 10

maximum in situation (A)

maximum in situation (B)

maximum in situation (C)

the same in all situations

Solution

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