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1

AIEEE 2012

MCQ (Single Correct Answer)
A coil is suspended in a uniform magnetic field, with the plane of the coil parallel to the magnetic lines of force. When a current is passed through the coil it starts oscillating; It is very difficult to stop. But if an aluminium plate is placed near to the coil, it stops. This is due to :
A
development of air current when the plate is placed
B
induction of electrical charge on the plate
C
shielding of magnetic lines of force as aluminium is a para-magnetic material.
D
electromagnetic induction in the aluminium plate giving rise to electromagnetic damping.

Explanation

Because of the Lenz's law of conservation of energy.
2

AIEEE 2011

MCQ (Single Correct Answer)
A resistor $$'R'$$ and $$2\mu F$$ capacitor in series is connected through a switch to $$200$$ $$V$$ direct supply. Across the capacitor is a neon bulb that lights up at $$120$$ $$V.$$ Calculate the value of $$R$$ to make the bulb light up $$5$$ $$s$$ after the switch has been closed. $$\left( {{{\log }_{10}}2.5 = 0.4} \right)$$
A
$$1.7 \times {10^5}\,\Omega $$
B
$$2.7 \times {10^6}\,\Omega $$
C
$$3.3 \times {10^7}\,\Omega $$
D
$$1.3 \times {10^4}\,\Omega $$

Explanation

We have, $$V = {V_0}\left( {1 - {e^{ - t/RC}}} \right)$$

$$ \Rightarrow 120 - 200\left( {1 - {e^{ - t/RC}}} \right)$$

$$ \Rightarrow t = RC\,in\,\left( {2.5} \right)$$

$$ \Rightarrow R = 2.71 \times {10^6}\Omega $$
3

AIEEE 2011

MCQ (Single Correct Answer)
A boat is moving due east in a region where the earth's magnetic fields is $$5.0 \times {10^{ - 5}}$$ $$N{A^{ - 1}}\,{m^{ - 1}}$$ due north and horizontal. The best carries a vertical aerial $$2$$ $$m$$ long. If the speed of the boat is $$1.50\,m{s^{ - 1}},$$ the magnitude of the induced $$emf$$ in the wire of aerial is :
A
$$0.75$$ $$mV$$
B
$$0.50$$ $$mV$$
C
$$0.15$$ $$mV$$
D
$$1$$ $$mV$$

Explanation

Induced $$emf$$ $$ = v{B_H}l = 1.5 \times 5 \times {10^{ - 5}} \times 2$$

$$ = 15 \times {10^{ - 5}} = 0.15\,mV$$
4

AIEEE 2011

MCQ (Single Correct Answer)
A fully charged capacitor $$C$$ with initial charge $${q_0}$$ is connected to a coil of self inductance $$L$$ at $$t=0.$$ The time at which the energy is stored equally between the electric and the magnetic fields is :
A
$${\pi \over 4}\sqrt {LC} $$
B
$$2\pi \sqrt {LC} $$
C
$$\sqrt {LC} $$
D
$$\pi \sqrt {LC} $$

Explanation

Energy stored in magnetic field $$ = {1 \over 2}L{i^2}$$

Energy stored in electric field $$ = {1 \over 2}{{{q^2}} \over C}$$

$$\therefore$$ $${1 \over 2}L{i^2} = {1 \over 2}{{{q^2}} \over C}$$

Also $$q = {q_0}\,\cos \,\omega t$$ and $$\omega = {1 \over {\sqrt {LC} }}$$

On solving $$t = {\pi \over 4}\sqrt {LC} $$

Questions Asked from Alternating Current and Electromagnetic Induction

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