Electromagnetic Induction Test

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Electromagnetic Induction Test - 23

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Question 1
1 / -0

Given below are the symbols of a few electronic components. Which of these components denote a variable inductor ?

A

B

C

D

Solution

An inductor whose inductance can be adjusted is known as variable inductor. It may consist, for instance, of a coil whose inductance can be varied by using one or more moving elements such as taps or sliding contacts. This contrasts with a fixed inductor, which has a single inductance which cannot be varied.

The variable inductor is represented as shown in the figure.
Question 2
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In one second, a current of 10 A changes through a coil. The induced emf is 10V, then, self-inductance of the coil is

A
2H

B
5H

C
1H

D
0.5H

Solution

From Faradays law the induced emf is$$e= L \frac {di}{dt}$$
Given, $$di = 10A, dt= 1s, e=10V$$
$$\therefore 10 = L\frac {10}{1}$$
$$\Rightarrow L = 1H$$
Question 3
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.................. gave the principle of Electromagnetic Induction.

A
Volta

B
Ampere

C
Faraday

D
Orested
Question 4
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If a magnet is plugged into a coil, then the magnitude of induced emf does not depend upon:

A
the number of turns in the coil

B
the medium of the core of the coil

C
the insertion speed of the magnet

D
the strength of the magnet

E
the resistance of the coil

Solution

The magnitude of induced emf is given by $$e=\displaystyle\frac{Nd\phi}{dt}=NB\displaystyle\frac{ds}{dt}$$
Which depends upon number of turns(N), magnetic field intensity(B) which depends upon medium, rate of change of area$$(ds/dt)$$ which depends upon insertion speed of magnet. It is independent of resistance of coil.
Question 5
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The self-inductance of a long solenoid carrying current is independent of:

A
its length

B
the current

C
its cross-sectional area

D
magnetic permeability of the core

E
the number of turns

Solution

Self-inductance of a solenoid is given by $$L=\displaystyle\frac{\mu_0N^2A}{l}$$
where, $$N=$$ number of turns in solenoid
$$l=$$ length of the solenoid
$$A=$$ area of one turn of solenoid.
The self-inductance of solenoid does not depend on current passing through it.
Question 6
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Which of the following does not have the same dimensions as the Henry?

A
$$\dfrac{\text {joule}}{(\text{ampere})^2}$$

B
$$\dfrac{\text {tesla} - m^2}{(\text{ampere})^2}$$

C
$$\text{ohm-second}$$

D
$$\dfrac{1}{\text{Farad-second}}$$

Solution

Option $$D$$ is does not have the same dimensions as the Henry.
So, Option $$D$$ is correct.
Question 7
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Which device is used to produce electricity? Describe with a neat diagram.

A
Electric motor

B
Galvanometer

C
Electric Generator (DC)

D
Voltmeter

Solution

The device used for producing electricity is Electric generator $$(DC)$$. It is based on the phenomenon of electromagnetic induction.
Question 8
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The current in a coil of 0.4 mH coil increases by 250 mA in 0.1 seconds. The induced emf will be:-

A
+1 V

B
-1 V

C
+ 1 mV

D
-1 mV

Solution

Fig: Inductor circuit

The emf induced in inductor i,
$$\varepsilon= -L\dfrac{dI}{dt}$$
Given,
$$L=0.4mH$$
$$dI=250mA$$
$$dt=0.1sec$$
$$\varepsilon=-\dfrac{0.4\times 10^{-3}\times 250\times10^{-3}}{0.1}$$
$$\varepsilon=-1mV$$
The correct Option is $$D$$
Question 9
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A solenoid of inductance $$L$$ carrying a certain current is linked with a total magnetic flux $$\phi $$. Now it is connected to a condenser with which it shares half of its initial energy. The total flux now linked with the solenoid is:

A
$$\dfrac{\phi}{2}$$

B
$$\dfrac{\phi}{\sqrt{2}}$$

C
$$\dfrac{\phi}{2\sqrt{2}}$$

D
$$\dfrac{\phi}{4}$$

Solution

Energy in indicator $$= \dfrac { 1 }{ 2 } L{ i }^{ 2 }$$
With Energy transfer, $$i$$ becomes $${ i }/{ \sqrt { 2 } }$$
Now since $$\phi = Li $$
New $$\phi ={ Li }/{ \sqrt { 2 } }={\phi}/{ \sqrt { 2 } }$$
So, total flux will be $${ \phi }/{ \sqrt { 2 } }$$
Question 10
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A coil with $$100$$ turns has an inductance of $$0.05H$$ and $$0.02A$$ current is passed through it. Flux linked with the coil is

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

B
$$10^{-3}Wb$$

C
$$10^{-4}Wb$$

D
$$10^{-5}Wb$$

Solution

As we know that flux = inductance x current
Hence,
Flux $$\phi = {Li} $$
$$\displaystyle = {0.05 \times 0.02}$$
$$=10^{-3} Wb $$