Electromagnetic Induction Test

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Electromagnetic Induction Test - 40

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Question 1
1 / -0

The magnetic flux through a circuit of resistance 'R' changes by an amount $$\Delta \Phi$$ at a time $$\Delta t$$. Then the total quantity of electric charge Q that passes any point in the circuit during the time $$\Delta t $$ is represented by

A
$$Q = \displaystyle \frac{\Delta \Phi}{R}$$

B
$$Q= \displaystyle \frac{\Delta \Phi}{\Delta t}$$

C
$$Q= \displaystyle R. \frac{\Delta \Phi}{\Delta t}$$

D
$$Q = \displaystyle {1}{R} . \frac{\Delta \Phi}{\Delta t}$$

Solution

$$\displaystyle I = \frac{e}{R} = \frac{1}{R} \left | \frac{\Delta \phi}{\Delta t} \right | = \frac{\Delta Q}{\Delta t}$$
$$\displaystyle Q = \frac{\Delta \phi}{R}$$
Question 2
1 / -0

Comment on the statement given below:
In self-induction
When the current in a coil is increasing, induced emf opposes it
When the current in a coil is decreasing, induced emf supports it

A
A is true, B is false

B
A and B are false

C
A and B are true

D
A is false, B is true

Solution

In self-induction,induced emf always opposes when the current in a coil is changing either increasing or decreasing.
So, A is true, B is false because when the current in a coil is increasing or decreasing, induced emf opposes it .
Therefore, A is correct option.
Question 3
1 / -0

When a rectangular coil is rotated in a uniform magnetic field about an axis passing through its centre and perpendicular to the field, the induced emf in the coil is maximum when the plane of the coil is

A
Perpendicular to the field

B
Parallel to the field

C
Inclined at $$60$$ with the field direction

D
Inclined at $$45$$ with the field direction

Solution

According to Faraday's law when the magnetic flux linking a circuit changes, an electromotive force is induced in the circuit proportional to the rate of change of the flux linkage.
So, when rectangular coil is rotated in a uniform magnetic field about an axis passing through its centre and perpendicular to the direction of the field, then the induced voltage in the coil $$=\dfrac{\mathrm{d} \phi}{\mathrm{d} t}$$
$$\Rightarrow e.m.f = $$$$\dfrac{\mathrm{d}( NAB\sin\theta)}{\mathrm{d} t}$$
$$\Rightarrow e.m.f = $$ $$NAB\omega\cos\theta$$
where $$\theta $$ is the the angle between the plane of the coil and the lines of force.
So, the induced emf in the coil is maximum when $$\cos\theta$$ is maximum or $$\theta$$ is zero means when the plane of the coil is`parallel to the field.
Therefore, B is correct option.
Question 4
1 / -0

A wire 88 cm long bent into a circular loop is placed perpendicular to the magnetic field of flux density 2.5 Wb $$m^{-2}$$. Within 0.5 s, the loop is changed into a square and flux density is increased to 3.0 Wb $$m^{-2}$$. The value of e.m.f. induced is :

A
0.018V

B
0.016V

C
0.020V

D
0.012V

Solution

Answer is A.

The magnetic flux $$\phi =B\times A$$ depends on the magnetic field, the area
of the loop, and their relative orientation.
In this case, the wire 88 cm long bent into a circular loop is placed perpendicular to the magnetic field and Within 0.5 s, the loop is changed into a square.
Magnetic Flux when in a circular loop:
Circumference of the loop = $$2\times 3.14\times r\quad =\quad 88\quad cm$$.
So, radius of the loop r = 14 cm.
Magnetic field $$\phi =B\times A\quad =2.5\times 3.14\times 0.14\times 0.14=0.154\quad Wb$$.
Magnetic Flux when in a square:
Circumference of the square = $$4r\quad =\quad 88\quad cm$$
So, length of the side of the square is $$\frac { 88 }{ 4 } =\quad 22cm$$.
Magnetic feild $$\phi '=B\times A\quad =3\times 0.22\times 0.22=0.1452\quad Wb$$.
The total induced emf is $$\frac { 0.154-0.1452 }{ 0.5 } =0.018\quad V$$.
Therefore, the value of e.m.f induced is 0.018 V.
Question 5
1 / -0

A copper ring is moved towards the north pole of a bar magnet. Then

A
the ring will not be affected

B
the ring all tend to get warm

C
an alternating current will flow in the ring

D
the ring will be positively charged

Solution

When the copper ring is moved towards the north pole of the magnet, induced emf is generated in the ring according to the electromagnetic laws of induction. Now current in the ring due to induced emf will generate heat in the ring according to the Joule's law. Thus the copper ring tends to get warm.
Question 6
1 / -0

A metal ring with a cut as shown in the figure is held horizontally and a bar magnet is dropped though the ring, then the acceleration of the magnet is ________.

A
Less than $$g$$

B
Greater than $$g$$

C
Equal to $$g$$

D
Either $$(1)$$ or $$(2)$$

Solution

A metal ring with a cut as shown in the figure is held horizontally and a bar magnet is dropped through the ring since the ring is cut and emf is not generated because Induced emf is generated only in a closed circuit As faraday's law of Electromagnetic induction states First Law of Faraday's Electromagnetic Induction state that whenever a conductor is placed in a varying magnetic field emf are induced which is called induced emf, if the conductor circuit are closed current are also induced which is called induced current. Hence the acceleration of the magnet is equal to $$g$$
Question 7
1 / -0

Faraday's second law of electromagnetic induction states that:

A
The magnitude of the induced emf is directly proportional to the rate of change of flux

B
The magnitude of the induced emf is inversely proportional to the rate of change of flux

C
The direction of the induced emf is such that it opposes the change in flux

D
The magnitude of the induced emf is directly proportional to the square of the rate of change of flux

Solution

Faraday's second law of electromagnetic induction states that The magnitude of the induced emf is directly proportional to the rate of change of flux.
Faraday's second law of electromagnetic induction states that, the magnitude of induced emf is equal to the rate of change of flux linkages with the coil. The flux linkages is the product of number of turns and the flux associated with the coil.
Question 8
1 / -0

The current flowing through the primary coil of mutual inductance 8 H is reduced to zero in $$10^{-3}$$s. As a result of which a $$24\times 10^3$$ V is induced in the secondary. The initial current through the primary is _____A.

A
2

B
1

C
4

D
3

Solution

Given,

$$M= 8 H$$

$$dI= I_2-I_1= -I_1A$$

$$dt= 10^{-3}s$$

$$\varepsilon= 24\times10^3 V$$

The induced emf $$\varepsilon$$ in the secondary coil is,

$$\varepsilon= -M\dfrac{dI}{dt}$$

$$24\times 10^3=\dfrac{8\times I_1}{10^{-3}}$$

$$I_1= \dfrac{24}{8}A$$

$$I_1=3A$$
Question 9
1 / -0

What will be the magnitude of e.m.f. induced in a 200 turns coil with cross section area 0.16 $$m^2$$ if the magnetic field through the coil changes from 0.10 Wb $$m^{-2}$$ to 0.30 $$m^{-2}$$ in 0.05 second.

A
128V

B
130V

C
118V

D
132V

Solution

Answer is A.

The magnetic flux $$\phi =B\times A$$ depends on the magnetic field, the area of the loop, and their relative orientation.
In this case, The magnetic field through the coil changes from $$0.10Wb{m}^{-2}$$ to $$0.30Wb{m}^{-2}$$, at a uniform rate over a period of 0.05 s.
The change in magnetic feild is $$0.10Wb{m}^{-2}$$ - $$0.30Wb{m}^{-2}$$ = $$0.20Wb{m}^{-2}$$.
The magnetic flux in one turn of the coil = $$\phi =B\times A\quad =\quad 0.20\times 0.16\quad =\quad 0.032T$$
Therefore, the Magnetic flux in 200 turns of coil = 0.032 * 200 = 6.4 T.
Therefore, the e.m.f induced in it, if this coil is removed from the field in 0.05 second is d$$\phi $$/dt = 6.4/0.05 = 128 V.
The magnitude of the e.m.f induced is 128 V.
Question 10
1 / -0

The variation of anode current in a triode valve corresponding to a change in grid potential at three different values of the plate potential is shown in the given figure. The mutual conductance of triode is

A
$$5\times {10}^{-3} mho$$

B
$$2.5\times {10}^{-3} mho$$

C
$$7.5\times {10}^{-3} mho$$

D
$$9.5\times {10}^{-3} mho$$

Solution

Mutual conductance

$$=\cfrac{\Delta {I}_{p}}{\Delta {V}_{g}}=\cfrac{5\times {10}^{-3}}{2}$$

$$=2.5\times {10}^{-3}mho$$

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Electromagnetic Induction Test - 40

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Electromagnetic Induction Test - 40

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Question 1

$$Q = \displaystyle \frac{\Delta \Phi}{R}$$

$$Q= \displaystyle \frac{\Delta \Phi}{\Delta t}$$

$$Q= \displaystyle R. \frac{\Delta \Phi}{\Delta t}$$

$$Q = \displaystyle {1}{R} . \frac{\Delta \Phi}{\Delta t}$$

Solution

Question 2

A is true, B is false

A and B are false

A and B are true

A is false, B is true

Solution

Question 3

Perpendicular to the field

Parallel to the field

Inclined at $$60$$ with the field direction

Inclined at $$45$$ with the field direction

Solution

Question 4

0.018V

0.016V

0.020V

0.012V

Solution

Question 5

the ring will not be affected

the ring all tend to get warm

an alternating current will flow in the ring

the ring will be positively charged

Solution

Question 6

Less than $$g$$

Greater than $$g$$

Equal to $$g$$

Either $$(1)$$ or $$(2)$$

Solution

Question 7

The magnitude of the induced emf is directly proportional to the rate of change of flux

The magnitude of the induced emf is inversely proportional to the rate of change of flux

The direction of the induced emf is such that it opposes the change in flux

The magnitude of the induced emf is directly proportional to the square of the rate of change of flux

Solution

Question 8

2

1

4

3

Solution

Question 9

128V

130V

118V

132V

Solution

Question 10

$$5\times {10}^{-3} mho$$

$$2.5\times {10}^{-3} mho$$

$$7.5\times {10}^{-3} mho$$

$$9.5\times {10}^{-3} mho$$

Solution

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