Electromagnetic Induction Test

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Electromagnetic Induction Test - 32

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Question 1
1 / -0

An athlete with 3m long iron rod in hand runs towards east with a speed of 30 kmph. The horizontal component of earth's magnetic field is $$4 \times 10^{-5} Wb/m^2$$. If he runs with the rod in horizontal and vertical positions then the induced emf generated in the rod in two cases will be

A
zero in vertical position and $$1 \times 10^{-3}$$ volt in horizontal position.

B
$$1 \times 10$$ volt in vertical position and zero volt in horizontal position.

C
zero in both positions

D
$$1 \times 10^{-3}$$ volt in both positions

Solution

In both the cases, the velocity vector is $$ \vec v = 30km/hr $$ towards east
Magnetic field = $$ \vec B = 4 \times 10 ^{-5} Wb/m^2 $$ towards north.

Force on a charge in the rod = $$ F = q \vec v \times \vec B = \dfrac{q}{3000} N $$ towards vertically down direction.

Therefore, when the rod is horzontal no emf is induced between the two ends.
When the rod is vertical, emf = $$ B l v = 1 mV $$ is induced between the two ends.
Question 2
1 / -0

When a coil of cross-sectional area $$A$$ and number of turns $$N$$ is rotated in a uniform magnetic field $$B$$ with angular velocity $$\omega$$, then the maximum emf induced in the coil will be

A
$$BNA$$

B
$$\displaystyle \frac{B A \omega}{N}$$

C
$$BNA$$ $$\omega$$

D
$$zero$$

Solution

$$emf =$$ rate of change of flux
$$=BAN\omega$$
Question 3
1 / -0

The resistance coils in a resistance box are made of double folded wire so that their

A
self induction effect in nullified

B
self inductance is maximum

C
induced emf is maximum

D
None of these

Solution

Presence of inductance increases the impedance more than the actual resistance value. Therefore to reduce the chance of occurrence of inductance in the resistor coils, they are double folded, so that current in opposite direction cancel the induced flux linkages if any and hence inductance is nullified.
Question 4
1 / -0

A conducting rod of length L is falling with velocity V in a uniform horizontal magnetic field B normal to the rod. The induced emf between the ends the rod will be :

A
2 BV$$l$$

B
zero

C
B$$l$$V

D
$$\displaystyle \frac{BVl}{2}$$

Solution

Force on charge q due to the motion of rod in the field= $$ F = q V B $$
This force on the charge is attributed to the induced electric field $$E$$
Therefore,
$$ Eq = F = q V B $$
$$ E = V B $$
therefore, an electric field $$ E = V B $$ is said to be induced in the rod due to the motion in the magnetic field.
Potential difference due to the field = emf induced =$$ E \times length = E l = VBl volts $$
Question 5
1 / -0

A millivoltmeter is connected in parallel to an axle of the train running with a speed of 180 km/hour. If the vertical component of earth's magnetic field is $$0.2 \times 10^{-4} Wb/m^2$$ and the distance between the rails is 1m, then the reading of voltmeter will be :

A
10$$^{-2}$$ volt

B
10$$^{-4}$$ volt

C
10$$^{-3}$$ volt

D
1 volt

Solution

Reading of voltmeter= emf induced = $$ B l v = 0.2 \times 10^{-4} \times 1 \times 180 \times \dfrac{1000}{3600}=1 mV $$
Question 6
1 / -0

The coefficients of self induction of two coils are $$L_1 = $$ 8mH and $$L_2=$$ 2mH respectively. The current rises in the two coils at the same rate. The power given to the two coils at any instant is same. The ratio of induced emf's in the coils will be :

A
$$\displaystyle \frac{V_1}{V_2} = 4$$

B
$$\displaystyle \frac{V_1}{V_2} = \frac{1}{4}$$

C
$$\displaystyle \frac{V_1}{V_2} = \frac{1}{2}$$

D
$$\displaystyle \frac{V_1}{V_2} = \frac{1}{3}$$

Solution

Power to 1st coil = $$ V_1 i_1 = L_1 \dfrac{di_1}{dt} i_i $$
Power to 2nd coil = $$ V_2 i_2 = L_2 \dfrac{di_2}{dt} i_2 $$
As the powers are same,
$$  L_1 \dfrac{di_1}{dt} i_i =  L_2 \dfrac{di_2}{dt} i_2 $$
rate of change of current is given to be same,
$$  L_1 i_i = L_2 i_2 $$
or, $$ \dfrac {i_1}{i_2} = \dfrac{ L_2}{L_1} =\dfrac{ 1}{4} $$
So, $$ \dfrac{ V_1}{V_2} = \dfrac{  L_1 \dfrac{di_1}{dt} } {  L_2 \dfrac{di_2}{dt} } = \dfrac{ L_1}{L_2} = 4 $$
Question 7
1 / -0

The coefficient of mutual inductance of the two coils is $$5H$$. The current through the primary coil is reduced to zero value from $$3 A$$ in $$1 $$millisecond. The induced emf in the secondary coils is

A
$$zero$$

B
$$1.67 KV$$

C
$$15 KV$$

D
$$600 V$$

Solution

Flux linking the secondary coil due to current in primary $$= M i_1$$

induced emf in the secondary coils $$=$$ rate of change of flux

$$ =M \dfrac{\triangle i_1}{\triangle t} $$

$$= 5 \times \dfrac{ 3-0}{0.001}$$

$$=15 k V$$
Question 8
1 / -0

The number of turns in a coil of wire of fixed radius is 600 and its self inductance is 108 mH. The self inductance of a coil of 500 turns will be :

A
74 mH

B
75 mH

C
76 mH

D
77 mH

Solution

$$L = \dfrac{\mu n^2 A}{L}$$.

$$L \propto n^2$$

Thus, for 500 turns, self inductance would be $$108\times (500)^2/(600)^2=108\times 25/36=75mH$$
Question 9
1 / -0

The magnetic flux linked with a coil is $$\phi \leq 8t^2 +3t + 5$$ Weber. The induced emf in fourth second will be

A
$$16 V$$

B
$$139 V$$

C
$$67 V$$

D
$$145 V$$

Solution

emf(t) = $$ \dfrac{ d \phi}{dt} = 16t+3 $$

emf(4)= $$16 \times 4 +3 = 67 V $$
Question 10
1 / -0

The distance between the ends of the wings of an airplane is $$50 m$$. It is flying in a horizontal plane at a speed of $$360 Km/hour$$. The vertical component of earth's magnetic field at that place is $$2.0 \times 10^{-4} Wb/m^2$$, then the potential difference induced between the ends of the wings will be

A
$$0.1 volt$$

B
$$1.0 volt$$

C
$$0.2 volt$$

D
$$0.01 volt$$

Solution

$$emf = B l v $$

$$= 2 \times 10^{-4} \times 50 \times 360 \times \dfrac{ 1000}{3600} = 1 V $$

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Electromagnetic Induction Test - 32

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Electromagnetic Induction Test - 32

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Question 1

zero in vertical position and $$1 \times 10^{-3}$$ volt in horizontal position.

$$1 \times 10$$ volt in vertical position and zero volt in horizontal position.

zero in both positions

$$1 \times 10^{-3}$$ volt in both positions

Solution

Question 2

$$BNA$$

$$\displaystyle \frac{B A \omega}{N}$$

$$BNA$$ $$\omega$$

$$zero$$

Solution

Question 3

self induction effect in nullified

self inductance is maximum

induced emf is maximum

None of these

Solution

Question 4

2 BV$$l$$

zero

B$$l$$V

$$\displaystyle \frac{BVl}{2}$$

Solution

Question 5

10$$^{-2}$$ volt

10$$^{-4}$$ volt

10$$^{-3}$$ volt

1 volt

Solution

Question 6

$$\displaystyle \frac{V_1}{V_2} = 4$$

$$\displaystyle \frac{V_1}{V_2} = \frac{1}{4}$$

$$\displaystyle \frac{V_1}{V_2} = \frac{1}{2}$$

$$\displaystyle \frac{V_1}{V_2} = \frac{1}{3}$$

Solution

Question 7

$$zero$$

$$1.67 KV$$

$$15 KV$$

$$600 V$$

Solution

Question 8

74 mH

75 mH

76 mH

77 mH

Solution

Question 9

$$16 V$$

$$139 V$$

$$67 V$$

$$145 V$$

Solution

Question 10

$$0.1 volt$$

$$1.0 volt$$

$$0.2 volt$$

$$0.01 volt$$

Solution

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