Electromagnetic Induction Test

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Electromagnetic Induction Test - 33

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Question 1
1 / -0

The value of mutual inductance can be increased by

A
decreasing N

B
increasing N

C
winding the coil on wooden frame

D
winding the coil on china clay

Solution

$$M = \mu N_1 N_2 \dfrac{ Area_{12}}{length_{12}} $$

$$ Area_{12} $$ - area in common to both the coils where the flux links both of them together.
$$length_{12}$$ - length in common to both the coils where the flux links both of them together.
Therefore to increase the mutual inductance, the number of turns can be increased.
Question 2
1 / -0

The value of coefficient of mutual induction for the arrangement of two coils shown in the figure will be :

A
Zero

B
Maximum

C
Negative

D
Positive

Solution

The mutual inductance between two coils depends upon the manner, in which, two coils are placed relative to each other. In the given figure, the magnetic flux linked with a coil due to current in another coil, seems to be zero, therefore coefficient of mutual inductance will be zero.
Question 3
1 / -0

The magnetic flux in a coil of 100 turns increases by $$12 \times 10^3$$ Maxwell in 0.2 second due to the motion of a magnet. The emf induced in the coil will be

A
6 V

B
0.6 V

C
0.06 V

D
60 V

Solution

$$emf =$$ rate of change of flux $$\times $$ number of turns $$= \dfrac{12 \times 10^3 \times 10^{-8}}{0.2} \times 100 = 0.06 V $$

1 Maxwell = $$ 10^{-8} $$ weber
Question 4
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A circular coil of conducting wire has an area A and number of turns N. It is lying in a vertical plane in a region where uniform magnetic field B exist with field direction normal to the coil plane. If the coil is rotated about a vertical axis by an angle $$\pi$$ in 0.5 seconds, then the value of the emf induced at the ends of the coil is

A
4 NAB

B
4$$\pi$$ NAB

C
8 NAB

D
8$$\pi$$ NAB

Solution

AS the coil rotates by $$\pi$$ radians, $$ \triangle \phi = NAB - (-NAB)=2NAB$$

$$emf =$$ rate of change of flux

$$=\dfrac{ \triangle \phi }{\triangle t } = \dfrac{ 2 NAB}{0.5}$$

$$ =4 NAB $$
Question 5
1 / -0

The two rails of a railways track, insulated from each other and the ground, are connected to a milli voltmeter. What is the reading of the milli voltmeter when a train travels at a speed of 20 ms$$^{-1}$$ along the track, given that the vertical component of the earth's magnetic field is 0.2 $$\times$$10$$^{-4}$$ Wbm$$^{-2}$$ and the rails are separated by 1 m?

A
4 mV

B
0.4 mV

C
80 mV

D
10 mV

Solution

emf = $$ B l v $$
$$=0.2 \times 10 ^{-4} \times 1 \times 20 $$
$$= 0.4 m V $$
Question 6
1 / -0

Flux $$\varphi$$ (in water) in a closed circuit of resistance 10 $$\Omega$$ varies with time t (in sec) according to the equation $$\varphi = 6t^2 - 5t + 1$$. What is the magnitude of the induced current at $$t = 0.25$$s?

A
1.2 A

B
0.8 A

C
0.6 A

D
0.2 A

Solution

$$emf = -\dfrac{d \phi}{dt}= -12t + 5 $$

current = $$- \dfrac{emf}{10} = -1.2 t + 0.5 $$

current at $$t= 0.25 s$$,

$$=-1.2 \times 0.25 + 0.5 =0.2 A$$
Question 7
1 / -0

Four pieces each of length $$l$$ of a conducting rod are joined as shown in the figure. The rod is placed in a downward magnetic field B. If the rod is moved towards right with a uniform velocity v, then the induced emf across the two ends of the rod is

A
Blv

B
4 Blv

C
2 Blv sin $$\left ( \theta/2 \right )$$

D
2 Blv $$\left (1 + sin \theta/2 \right )$$

Solution

From figure $$b = lsin(\theta/2)$$

Length of rod $$L = l+2b+l = 2l+2lsin(\theta/2) = 2l[1+sin(\theta/2)]$$

Induced emf between $$A $$ and $$B$$ $$\mathcal{E} = BvL$$

$$\implies$$ $$\mathcal{E} = 2lBv[1+sin(\theta/2)]$$
Question 8
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A conducting rod of length L is falling with velocity v perpendicular to a uniform horizontal magnetic field B. The potential difference between its two ends will be

A
2BLv

B
BLv

C
$$\frac{1}{2}$$BLv

D
(BLv)$$^2$$

Solution

Force on a charge q in the rod = $$ q v B $$

electric field inside the rod due to displacements of charges due to force by relative motion in magnetic field $$= \dfrac{emf}{l} $$

As the rod moves in constant velocity, net force of constituent charge q in the rod $$=0$$
Therefore,

$$ q\dfrac{emf}{l} = q v B$$

$$ \Rightarrow emf = B l v$$
Question 9
1 / -0

Directions For Questions

A captain is shipwrecked on a deserted tropical island. He possesses some electrical devices that could be operated using a generator but he does not have magnets. The earth's magnetic field at that location is horizontal and is equal to $$\displaystyle 8 \times 10^{-5} T.$$. He decides to use this magnetic field for a generator by rotating a large circular coil of wire at a high rate. His requirement is $$9\ V$$ and he thinks he can rotate the coil at a speed $$30\ rpm$$ by turning a crank handle. The number of turns in the coil are $$2000$$.

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The area of the coil must be._____

A
$$1.8\ m^2$$

B
$$18\ m^2$$

C
$$8\ m^2$$

D
$$none\ of\ these$$

Solution

Flux through a circular coil $$\phi=NBA\cos \omega t$$

Voltage required $$\displaystyle \varepsilon = \dfrac{-d \phi}{dt} $$

$$\Rightarrow 9 = NBA \omega sin \omega t$$

$$\displaystyle 9 = \dfrac{8 \times 10^{-5} \times A \times 30 \times 2\pi \times 2000}{60}$$

$$\displaystyle A = \dfrac{9 \times 10^5}{50 \times 10^3} = 18m^2$$
Question 10
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A conducting wire in the shape of Y, with each side of length $$l$$ is moving in a uniform magnetic field B, with a uniform speed v as shown in the figure. The induced emf at the two ends X and Y of the wire will be

A
zero

B
2Blv

C
2Blv sin($$\theta$$/2)

D
2Blv cos($$\theta$$/2)

Solution

From figure $$b = lsin(\theta/2)$$

Emf induced between X and Y

$$\mathcal{E_{xy}} = BvL$$

where $$L = 2b$$

$$\therefore$$ $$\mathcal{E_{xy}} =Bv(2b)= 2Blv lsin(\theta/2)$$