Electromagnetic Induction Test

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Electromagnetic Induction Test - 35

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Question 1
1 / -0

Two coils, A and B, are lined such that emf $$\epsilon$$ is induced in B when the current in A is changing at the rate $$I$$. If $$i$$ current is now made to flow in B, the flux linked with A will be :

A
$$(\epsilon / I)i$$

B
$$\epsilon i I$$

C
$$(\epsilon I)i$$

D
$$i I/\epsilon$$

Solution

Let $$M=$$ mutual inductance between A and B.

$$\epsilon_B=M\dfrac {di_A}{dt}$$ or $$\epsilon_B=MI$$ or

$$M=\epsilon_B/I$$

where $$I=di_A/dt$$

$$\phi_A=Mi_B=(\epsilon_B/I)i_B$$

$$\therefore \phi_A=(\epsilon /I)i$$ $$ (\because i=i_B)$$

Question 2

1 / -0

Coils in the resistance boxes are made from doubled up insulated wires

to cancel the effect of self induction

to nullify the heating effect

to nullify the Peltier effect

to reduce effective length of the wire

Solution

The wire is doubled back on itself. As a result ,there are equal and opposite currents in each section of the coil. Therefore ,the coil has no net magnetic field and no net induced e.m.f.
Thus, it is done to minimize the inductance of coils.

Question 3
1 / -0

If a current increases from zero to one ampere in 0.1 second in a coil of 5 mH, then the magnitude of the induced e.m.f. will be

A
0.005 volt

B
0.5 volt

C
0.05 volt

D
5 volt

Solution

$$\varepsilon=L\times (rate \ of \ change \ of \ current)$$

$$=(5\times 10^{-3})(1/0.1)$$

$$=0.05V$$.
Question 4
1 / -0

The SI unit of inductance, the henry can be written as

A
weber/ ampere

B
volt second/ ampere

C
joule/ $$ampere^2$$

D
all of the above

Solution

$$ e = \dfrac{d\phi}{dt} = L\dfrac{dI}{dt} $$

$$\Rightarrow [volt] = \dfrac{[weber]}{[second]} = [L]\dfrac{[Ampere]}{[second]}$$

also,
Energy dissipated(E) $$ = VIt$$

$$\Rightarrow [Joule] = [volt][Ampere][second] \ Rightarrow [volt-second] = \dfrac{[Joule]}{[Ampere]}$$

so units of self inductance L is,

$$[L] = \dfrac{weber}{Ampere} = \dfrac{volt - second}{Ampere} = \dfrac{Joule}{Ampere^2}$$
Question 5
1 / -0

A copper ring is suspended by a thread in a vertical plane. If one end of a magnet is brought horizontally towards the ring in plane of ring as shown, the ring will

A
move towards the magnet.

B
not change its position.

C
move away from the magnet.

D
first move towards and then move away from the magnet

Solution

Answer is B.

The magnetic flux is given by $$\phi= BA cos \theta$$ where $$\theta$$ is angle between magnetic field and area of ring.
in this case the area vector is out of the plane, hence $$\theta=90^\circ{}$$,
$$\Rightarrow \phi =0$$
The ring will not move from the magnet, as there is no change in flux through ring and current in it will be zero. Ring will not behave as a magnet.
Question 6
1 / -0

A generator has an e.m.f. of 440 volt and internal resistance of 400 ohm. Its terminals are connected to a load of 4000 ohm. The voltage across the load is:

A
$$220 volt$$

B
$$440 volt$$

C
$$200 volt$$

D
$$400 volt$$

Solution

Total resistance of the circuit$$=4000+400=4400\ W$$
Current flowing $$i=\dfrac {V}{R}=\dfrac {440}{4400}=0.1\ A$$.
Voltage across load $$V_L = Ri=4000\times 0.1=400\ V$$.
Question 7
1 / -0

A rectangular coil of single turn, having area A, rotates in a uniform magnetic field B with an angular velocity $$\omega$$ about an axis perpendicular to the field. If initially the plane of the coil is perpendicular to the field, then the average induced emf when it has rotated through $$90^o$$ is:

A
$$\dfrac {\omega BA}{\pi}$$

B
$$\dfrac {\omega BA}{2\pi}$$

C
$$\dfrac {\omega BA}{4\pi}$$

D
$$\dfrac {2\omega BA}{\pi}$$

Solution

Initially flux, $$\phi=BA cos 0=BA$$
After rotating through an angle $$90^o$$.
Flux through the coil is zero.
So, $$\Delta \phi=BA$$

Angular speed $$=w$$, so, time period $$=\dfrac {2\pi}{\omega}=T$$

$$\dfrac {T}{4}$$ is time taken to rotate $$90^o$$.

So, $$\dfrac {\Delta \phi}{\Delta t}=\dfrac {BA}{T/4}=\dfrac {2BA\omega}{\pi}$$
Question 8
1 / -0

A coil has 200 turns and area of $$70 cm^2$$. The magnetic field perpendicular to the plane of the coil is $$0.3 Wb/m^2$$ and take 0.1 sec to rotate through $$180^o$$. The value of the induced e.m.f. will be

A
8.4 V

B
84 V

C
42 V

D
4.2 V

Solution

Change in flux $$=2BAN$$

$$\therefore$$ Induced e.m.f. $$=\dfrac {2\times 0.3\times 200\times 70\times 10^{-4}}{0.1}$$

$$=8.4V$$
Question 9
1 / -0

According to Faraday's law of electromagnetic induction:

A
electric field is produced by time varying magnetic flux

B
magnetic field is produced by time varying electric flux

C
magnetic field is associated with a moving charge

D
none of these

Solution

According to Faraday's Law of electromagnetic induction, if magnetic flux through a coil changes, it leads to an induced emf across the coil. Hence, electric field is produced by time-varying magnetic fields.
Question 10
1 / -0

If coefficient of self induction of a coil is 1 H, an e.m.f. of 1 V is induced, if

A
current flowing is 1 A

B
current variation rate is 1 $$As^{-1}$$

C
current of 1A flows for one sec

D
none of these

Solution

Induced EMF $$E =-L\dfrac{di}{dt}$$...(i)

Given $$E=1\ V$$, $$L=1 \ H$$

$$\Rightarrow \dfrac{di}{dt} =1 A/s$$

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Electromagnetic Induction Test - 35

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Electromagnetic Induction Test - 35

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Question 1

$$(\epsilon / I)i$$

$$\epsilon i I$$

$$(\epsilon I)i$$

$$i I/\epsilon$$

Solution

Question 2

to cancel the effect of self induction

to nullify the heating effect

to nullify the Peltier effect

to reduce effective length of the wire

Solution

Question 3

0.005 volt

0.5 volt

0.05 volt

5 volt

Solution

Question 4

weber/ ampere

volt second/ ampere

joule/ $$ampere^2$$

all of the above

Solution

Question 5

move towards the magnet.

not change its position.

move away from the magnet.

first move towards and then move away from the magnet

Solution

Question 6

$$220 volt$$

$$440 volt$$

$$200 volt$$

$$400 volt$$

Solution

Question 7

$$\dfrac {\omega BA}{\pi}$$

$$\dfrac {\omega BA}{2\pi}$$

$$\dfrac {\omega BA}{4\pi}$$

$$\dfrac {2\omega BA}{\pi}$$

Solution

Question 8

8.4 V

84 V

42 V

4.2 V

Solution

Question 9

electric field is produced by time varying magnetic flux

magnetic field is produced by time varying electric flux

magnetic field is associated with a moving charge

none of these

Solution

Question 10

current flowing is 1 A

current variation rate is 1 $$As^{-1}$$

current of 1A flows for one sec

none of these

Solution

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