Electromagnetic Induction Test

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Electromagnetic Induction Test - 46

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Question 1
1 / -0

Given figure is the symbol for which component in the electrical circuit?

A
Inductor

B
LED

C
Resistor

D
Battery

Solution

The given symbol is of $$inductor$$.

Answer-(A)
Question 2
1 / -0

An e.m.f. of 5 millivolt is induced in a coil when in a nearby placed another coil, the current changes by 5 ampere in 0.1 second. The coefficient of mutual induction between the two coils will be :

A
1 Henry

B
0.1 Henry

C
0.1 millihenry

D
0.001 millihenry

Solution

Emf induced   $$|\mathcal{E}| = 5mV = 5\times 10^{-3} \ V$$
Rate of change of current   $$\dfrac{di}{dt} = \dfrac{5}{0.1} = 50 \ A/s$$
Mutual inductance between the coil   $$M = \dfrac{|\mathcal{E}|}{\dfrac{di}{dt}}$$
$$\implies \ M = \dfrac{5\times 10^{-3}}{50} = 10^{-4} \ H = 0.1 \ mH$$
Question 3
1 / -0

When the speed at which a conductor is moved through a magnetic field is increased, the induced voltage

A
increases

B
decreases

C
remains constant

D
reaches zero

Solution

$$E=vBl$$ where $$v$$=speed of conductor

Hence, on increasing speed of conductor , induced voltage increases..

Answer-(A)
Question 4
1 / -0

What is the magnetomotive force (mmf) of a wire with 8 turns carrying three amperes of current?

A
2,400 At

B
240 At

C
24 At

D
2.4 At

Solution

Current flowing through the wire $$i = 3 \ A$$
Number of turns $$N = 8$$
Thus magneto motive force $$ = Ni = 8\times 3 = 24 \ At$$
Question 5
1 / -0

Reactance of a coil is $$157\Omega$$. On connecting the coil across a source of frequency $$ 100Hz$$, the current lags behind e.m.f. by $${ 45 }^{ o }$$. The inductance of the coil is _________.

A
$$0.25 H$$

B
$$0.5 H$$

C
$$4H$$

D
$$314 H$$

Solution

Since the phase angle is $$45^{\circ}$$,

$$\dfrac{X_L}{R}=tan\phi=tan45^{\circ}=1$$

$$\implies X_L=R$$

$$\implies \omega L=R$$

$$\implies 2\pi f L=R$$

$$\implies L=\dfrac{R}{2\pi f}$$

$$=\dfrac{157}{2\pi\times 100}H$$

$$=0.25H$$
Question 6
1 / -0

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of $$10^{-3}$$ Wb to link with A and a flux per turn of $$0.8 \times 10^{-3}$$ Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is :

A
5/4

B
1/1.6

C
1.6

D
1

Solution

Two coils A and B have 200 and 400 turns respectively. A current of 1 A in coil A causes a flux per turn of 10−3 Wb to link with A and a flux per turn of 0.8×10−3 Wb through B. The ratio of self-inductance of A and the mutual inductance of A and B is $$\dfrac{L_1}{L_2}=\dfrac{200*10^{-3}}{400*0.8*10^{-3}}=1/1.6$$
Question 7
1 / -0

Electromagnetic induction is not used in :

A
Transformer

B
Room heater

C
AC generator

D
Choke coil

Solution

Room heater works on the principle of Joule's heating effect of current. It states that heat is produced in the conductor (or resistor) by passing an electric current through it for some time. While transformer, AC generator, and choke coil, all work on the principle of Electromagnetic Induction.

Hence, the correct option is B
Question 8
1 / -0

If an emf of $$25 V$$ is induced when the current in the coil changes at the rate of $$100\ As^{-1}$$, then the coefficient of self induction of the coil is

A
$$0.3\ H$$

B
$$0.25\ H$$

C
$$2.5\ H$$

D
$$0.25\ mH$$

Solution

Rate of change of current, $$\dfrac{di}{dt} = 100$$ $$As^{-1}$$
Magnitude of emf induced in the coil, $$|\mathcal{E}| = 25$$ V
Using: $$|\mathcal{E}| = L\dfrac{di}{dt}$$ where $$L$$ is the coefficient of self inductance
$$\implies$$ $$25 = L\times 100$$
$$\implies$$ $$L = 0.25$$ $$H$$
Question 9
1 / -0

Electromagnetic induction is not used in :

A
Transformer

B
Room heater

C
AC generator

D
Choke coil

Solution

Electromagnetic induction is used in transformers, AC generators and choke coils. This phenomenon is responsible for induction of voltage in secondary coils in transformers and for induction of voltage due to change to flux through the coil in AC generators. Choke coils are inductors used to block higher-frequency alternating current (AC) in an electrical circuit, while passing lower-frequency or direct current (DC).
However a room heater works on the principle of power radiated by passing current though a resistor in accordance with the Ohm's Law.
Question 10
1 / -0

Electromagnetic induction is not used in

A
Transformer

B
Room heater

C
AC generator

D
Choke coil

Solution

Room heater works on the principle of Joule's heating effect of current that heat is produced in the conductor (or resistor) on passing an electric current through it for some time.
While transformer, AC generator, and choke coil work on the principle of Electromagnetic induction.

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Electromagnetic Induction Test - 46

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Electromagnetic Induction Test - 46

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SHARING IS CARING

Question 1

Inductor

LED

Resistor

Battery

Solution

Question 2

1 Henry

0.1 Henry

0.1 millihenry

0.001 millihenry

Solution

Question 3

increases

decreases

remains constant

reaches zero

Solution

Question 4

2,400 At

240 At

24 At

2.4 At

Solution

Question 5

$$0.25 H$$

$$0.5 H$$

$$4H$$

$$314 H$$

Solution

Question 6

5/4

1/1.6

1.6

1

Solution

Question 7

Transformer

Room heater

AC generator

Choke coil

Solution

Question 8

$$0.3\ H$$

$$0.25\ H$$

$$2.5\ H$$

$$0.25\ mH$$

Solution

Question 9

Transformer

Room heater

AC generator

Choke coil

Solution

Question 10

Transformer

Room heater

AC generator

Choke coil

Solution

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