Electromagnetic Induction Test

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Electromagnetic Induction Test - 49

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Question 1
1 / -0

The horizontal component of earth's magnetic field at a place is $$3\times 10^{-4}T$$ and the dip is $$\tan^{-1}\left (\dfrac {4}{3}\right )$$. A metal rod of length $$0.25\ m$$ is placed in $$N-S$$ direction and is moved at a constant speed of $$10\ cms^{-1}$$ towards the east. The emf induced in the rod will be

A
$$1\mu V$$

B
$$5\mu V$$

C
$$7\mu V$$

D
$$10\mu V$$

Solution

Rod is moving towards east, so induced emf across its end will be
$$e = B_{v}vl = (B_{H}\tan \theta)vl \left [\therefore \tan \theta = \dfrac {B_{V}}{B_{H}}\right ]$$
$$= 3\times 10^{-4} \times \dfrac {4}{3} \times 10\times 10^{-3}\times 0.25$$
$$= 10^{-5}V = 10\mu V$$.
Question 2
1 / -0

What should be the value of self inductance of an indicator that should be connected to $$220$$V, $$50$$Hz supply so that a maximum current of $$0.9$$A flows through it?

A
$$11$$H

B
$$2$$H

C
$$1.1$$H

D
$$5$$H

Solution

The emf across the inductor is given by:
$$|e|=\displaystyle\frac{Ldi}{dt}$$

$$220=L\times \frac{0.9}{1/50}$$

$$L=4.88=5$$H.
Question 3
1 / -0

In A.C generator increasing no. of turns in coil :

A
decreases the EMF

B
EMF remains same

C
increases the EMF

D
EMF becomes zero

Solution

We know that emf generated in an AC generator is given by

$$E=NBA\omega \cos\omega t$$

Hence,on increasing number of turns of coil, EMF increases.

Answer-(C)
Question 4
1 / -0

In alternative current generator, AC current reverses its direction :

A
20 times per second

B
50 times per second

C
once per second

D
twice per second

Solution

The frequency of an AC generator is 50 times per second.
Hence, the current reverses its direction 50 times.

Answer-(B)
Question 5
1 / -0

The part of the AC generator that passes the current from the coil to the external circuit is?

A
Field magnet

B
Split rings

C
Slip rings

D
Brushes
Solution
- The three phase AC generator has three equally spaced windings and three output voltages that are $$120^{0}$$ out of phase of one another.
- It is based on the principle of faraday law of electromagnetic induction
- Field magnet passes the current from the coil to the external circuit in AC generator.
Question 6
1 / -0

A conducting rod of Length $$L=0.1$$m is moving with a uniform speed $$v=0.2$$ m/s on conducting rails in a magnetic field $$B=0.5$$T as shown. On one side, the end of the rails is connected to a capacitor of capacitance $$C=20\mu F$$. Then the charges on the capacitor plates are.

A
$$q_A=0=q_B$$

B
$$q_A=+20\mu C$$ and $$q_B=-20\mu C$$

C
$$q_A=+0.2\mu C$$ and $$q_B=-0.2\mu C$$

D
$$q_A=-0.2\mu C$$ and $$q_B=+0.2\mu C$$

Solution

$$L=0.1m$$ , $$v=0.2 ms^{-1}$$ ,$$B=0.5T$$ ,$$C=20 \mu F$$
$$V=BvL$$
=$$0.5 \times 0.2 \times 0.1$$=$$0.01$$
$$q=cv=20\mu F\times 0.01=0.2\mu C$$
$$q_A=+0.2 \mu C$$
$$q_B=-0.2 \mu C$$
Question 7
1 / -0

The number of turns in the coil of an AC generator ar e$$100$$ and its cross-sectional area is $$2.5{m}^{2}$$. The coil is revolving in a uniform magnetic field of strength $$0.3T$$ with the uniform angular velocity of $$60rad/s$$. The value of maximum value produced is _____kV

A
$$1.25$$

B
$$4.50$$

C
$$6.75$$

D
$$2.25$$

Solution

The induced emf in a coil is given by
$$e=NBA\omega\sin { \omega t } $$

for maximum induced emf $${ e }_{ max }$$
$$\sin { \omega t } =1$$
So $${ e }_{ max }=NBA\omega $$

putting values
$${ e }_{ max }=100\times0.3\times2.5\times60$$
$${ e }_{ max }=4.5kV$$
Question 8
1 / -0

If given arrangement is moving towards left with speed v, then potential difference between B and D and current in the loop are respectively.

A
BvR and non-zero

B
2BvR and zero

C
4BvR and non-zero

D
4BvR and zero

Solution

Curve wire moving transnational can be equalize by straight wire jointed to its free end.
Then potential difference between B & d will be =
$$ { V }_{ d }-{ V }_{ b }=\quad BV\times 4R\\ { V }_{ d }{ -V }_{ b }=4BVR$$
The current in the loop will be zero because we know that net emf in any closed loop is zero.
So option (D) is the correct answer.
Question 9
1 / -0

A copper rod of length $$l$$ rotates about its end with angular velocity $$\omega$$ in a uniform magnetic field B. The emf developed between the ends of the rod if the field is normal to the plane of rotation is:

A
$$B\omega l^2$$

B
$$\dfrac{1}{2}B\omega l^2$$

C
2 $$B\omega l^2$$

D
$$\dfrac{1}{4}B\omega l^2$$

Solution

$$dE=B\omega xdx\\E=\int _{ 0 }^{ L }{ B\omega xdx } \\ \quad =B\omega\cfrac{x^2}{2}\\ \quad=\cfrac{1}{2}B \omega l^2$$
Question 10
1 / -0

A wheel with 10 metallic spokes each 0.5 m long rotated with a speed of 120 rpm in a plane normal to the horizontal component of earth's magnetic field $$B_h$$ at a place. If $$B_h$$ = 0.4 G at the place. What is the induced emf between the axle and the rim of the wheel? (1 G = $$10^4$$ T)

A
0 V

B
0.628 mV

C
0.628 $$\mu$$V

D
62.8 $$\mu$$V

Solution

Given,

A wheel with $$410$$ metallic spokes.

Length of metallic spoke $$=0.5\,m$$

Angular speed of the wheel $$=120\,rev/min$$

Earth's magnetic field, $$B=0.4\,G$$

The area covered by an angle $$4\theta$$,

$$A=\pi r^2\dfrac{\theta}{2\pi}=\dfrac{r^2}{2}\theta$$

The induces emf,

$$E=\dfrac{d\phi}{dt}=\dfrac{d(BA)}{dt}$$

$$=B\dfrac{r^2}{2}\dfrac{d\phi}{dt}=B\dfrac{r^2}{2}\omega$$

$$v=120\,rev/min=2\,rev/s$$

$$\omega=2\pi r=4\pi rad/s$$

$$r=0.5\,m$$

$$B=0.4\,G=0.4\times10^{-4}\,T$$

Therefore,

Induced emf is given by,

$$E=\dfrac12 Br^2\omega$$

$$E=\dfrac{2\pi\times 2\times 0.4\times 10^{-4}\times (0.5^2)}{2}$$

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

Each spoke will act as a parallel source of emf.

Hence, the emf will be $$0.628\,mV$$

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Electromagnetic Induction Test - 49

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Electromagnetic Induction Test - 49

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Question 1

$$1\mu V$$

$$5\mu V$$

$$7\mu V$$

$$10\mu V$$

Solution

Question 2

$$11$$H

$$2$$H

$$1.1$$H

$$5$$H

Solution

Question 3

decreases the EMF

EMF remains same

increases the EMF

EMF becomes zero

Solution

Question 4

20 times per second

50 times per second

once per second

twice per second

Solution

Question 5

Field magnet

Split rings

Slip rings

Brushes

Solution

Question 6

$$q_A=0=q_B$$

$$q_A=+20\mu C$$ and $$q_B=-20\mu C$$

$$q_A=+0.2\mu C$$ and $$q_B=-0.2\mu C$$

$$q_A=-0.2\mu C$$ and $$q_B=+0.2\mu C$$

Solution

Question 7

$$1.25$$

$$4.50$$

$$6.75$$

$$2.25$$

Solution

Question 8

BvR and non-zero

2BvR and zero

4BvR and non-zero

4BvR and zero

Solution

Question 9

$$B\omega l^2$$

$$\dfrac{1}{2}B\omega l^2$$

2 $$B\omega l^2$$

$$\dfrac{1}{4}B\omega l^2$$

Solution

Question 10

0 V

0.628 mV

0.628 $$\mu$$V

62.8 $$\mu$$V

Solution

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