Electromagnetic Induction Test

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Electromagnetic Induction Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

The current flowing in a copper voltameter is 3.2 A. The number of copper ions $$(Cu^{2+})$$ deposited at the cathode per minute is

A
$$6\times 10^{20}$$

B
$$1.5\times 10^{20}$$

C
$$3\times 10^{20}$$

D
$$
3\times 10^{20}$$

Solution

Charge supplied per minute = $$ 3.2 \times 60 = 192 C $$
Charge 2 e liberates one $$ Cu^{+2} $$
No of $$ Cu^{+2} $$ ion liberates by 192 C
$$ n= \dfrac{192}{2e} = \dfrac{192}{2\times 1.6\times 10^{-19}} = 6\times 10^{20} $$
Question 2
1 / -0

In a coil of self inductance $$0.5$$ henry, the current varies at a constant rate from zero to 10 amperes in 2 seconds. The e.m.f.generated in the coil is

A
$$10$$ volts

B
$$5$$ volts

C
$$2.5$$ volts

D
$$1.25$$ volts

Solution

$$\dfrac{\Delta i}{\Delta t} = \dfrac{10}{2} = 5 \,A / sec $$

$$\Rightarrow e = L \dfrac{\Delta i}{\Delta t} = 0.5 \times 5 = 2.5 \,volts$$
Question 3
1 / -0

Which of the following is wrong statement

A
An emf can be induced between the ends of a straight conductor by moving it through a uniform magnetic field

B
The self induced emf produced by changing current in a coil always tends to decrease the current

C
Inserting an iron core in a coil increases its coefficient of self induction

D
According to Lenz's law, the direction of the induced current is such that it opposes the flux change that causes it

Solution
Question 4
1 / -0

A coil of wire of a certain radius has $$600$$ turns and a self inductance of $$108$$ mH. The self inductance of a $$2^{nd}$$ similar coil of $$500$$ turns will be

A
$$74$$ mH

B
$$75$$ mH

C
$$76$$ mH

D
$$77$$ mH

Solution

$$\dfrac{L_{B }}{L_{A}} = \left ( \dfrac{n_{B}}{n_{A}} \right )^2 $$

$$\Rightarrow L_{B } = \left ( \dfrac{500}{600} \right)^2 \times 108 = 75 \,mH$$
Question 5
1 / -0

A wheel with ten metallic spokes each $$0.50\, m$$ long is rotated with a speed of $$120\, rev/min$$ in a plane normal to the earths magnetic field at the place. If the magnitude of the field is $$0.4$$ Gauss, the induced e.m.f.between the axle and the rim of the wheel is equal to

A
$$1.256 \times 10^{-3} \,V$$

B
$$6.28 \times 10^{-4} \,V$$

C
$$1.256 \times 10^{-4} \,V$$

D
$$6.28 \times 10^{-5} \,V$$

Solution
Question 6
1 / -0

In the figure, shown the magnetic induction at the center of their arc due to the current in the potion $$AB$$ the magnetic induction at $$O$$ due to the whole length of the conductor is

A
$$\dfrac {\mu_0i}{r}$$

B
$$\dfrac {\mu_0i}{2r}$$

C
$$\dfrac {\mu_0i}{4r}$$

D
$$Zero$$

Solution

The induction due to $$AB$$ and $$CD$$ will be zero. Hence the whole induction will be due to the semicircular part $$BC$$.
$$B=\dfrac{\mu_0i}{4r}$$
Question 7
1 / -0

In the figure shown, the magnetic induction at the center of their arc due to the current in the potion $$AB$$ will be

A
$$\dfrac {\mu_0i}{r}$$

B
$$\dfrac {\mu_0i}{2r}$$

C
$$\dfrac {\mu_0i}{4r}$$

D
$$Zero$$

Solution

The magnetic induction at $$O$$ due to the current in portion $$AB$$ will be zero because $$O$$ lies on $$AB$$ when extended.
Question 8
1 / -0

The electrochemical equivalent of a metal is $$3.3\times 10^{-7}$$ kg / coulomb. The mass of the metal liberated at the cathode when a 3 A current is passed for 2 second will be

A
$$19.8\times 10^{-7}kg$$

B
$$9.39\times 10^{-7}kg$$

C
$$6.6\times 10^{-7}kg$$

D
$$1.1\times 10^{-7}kg$$

Solution

It is known that $$ m = Z it$$
according to question:
$$ m = Z it = 3.3\times 10^{-7} \times 3 \times 2 = 19.8 \times 10 ^{-7} kg $$
Question 9
1 / -0

Find out the e.m.f. produced when the current changes from 0 to 1 A in 10 second, given $$L = 10 \,\mu H$$

A
$$1 \,V$$

B
$$1 \,\mu V$$

C
$$1 \,mV$$

D
$$0.1 \,V$$

Solution

Given:
$$L = 10 \,\mu H\\t=10A\\i_i=10A\\i_f=1A$$

$$| e | = L \dfrac{di}{dt} \Rightarrow | e | = 10 \times 10^{-6} \times \dfrac{1}{10} = 1 \,\mu V$$
Question 10
1 / -0

Plane of eddy currents makes an angle with the plane of magnetic lines of force equal to

A
$$40^\circ$$

B
$$0^\circ$$

C
$$90^\circ$$

D
$$180^\circ$$

Solution

Direction of eddy currents is given by Lenz’s rule.
Hence Option C

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Electromagnetic Induction Test - 61

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Electromagnetic Induction Test - 61

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Question 1

$$6\times 10^{20}$$

$$1.5\times 10^{20}$$

$$3\times 10^{20}$$

$$3\times 10^{20}$$

Solution

Question 2

$$10$$ volts

$$5$$ volts

$$2.5$$ volts

$$1.25$$ volts

Solution

Question 3

An emf can be induced between the ends of a straight conductor by moving it through a uniform magnetic field

The self induced emf produced by changing current in a coil always tends to decrease the current

Inserting an iron core in a coil increases its coefficient of self induction

According to Lenz's law, the direction of the induced current is such that it opposes the flux change that causes it

Solution

Question 4

$$74$$ mH

$$75$$ mH

$$76$$ mH

$$77$$ mH

Solution

Question 5

$$1.256 \times 10^{-3} \,V$$

$$6.28 \times 10^{-4} \,V$$

$$1.256 \times 10^{-4} \,V$$

$$6.28 \times 10^{-5} \,V$$

Solution

Question 6

$$\dfrac {\mu_0i}{r}$$

$$\dfrac {\mu_0i}{2r}$$

$$\dfrac {\mu_0i}{4r}$$

$$Zero$$

Solution

Question 7

$$\dfrac {\mu_0i}{r}$$

$$\dfrac {\mu_0i}{2r}$$

$$\dfrac {\mu_0i}{4r}$$

$$Zero$$

Solution

Question 8

$$19.8\times 10^{-7}kg$$

$$9.39\times 10^{-7}kg$$

$$6.6\times 10^{-7}kg$$

$$1.1\times 10^{-7}kg$$

Solution

Question 9

$$1 \,V$$

$$1 \,\mu V$$

$$1 \,mV$$

$$0.1 \,V$$

Solution

Question 10

$$40^\circ$$

$$0^\circ$$

$$90^\circ$$

$$180^\circ$$

Solution

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$$
3\times 10^{20}$$