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1

AIEEE 2004

MCQ (Single Correct Answer)
In a uniform magnetic field of induction $$B$$ a wire in the form of a semicircle of radius $$r$$ rotates about the diameter of the circle with an angular frequency $$\omega .$$ The axis of rotation is perpendicular to the field. If the total resistance of the circuit is $$R,$$ the mean power generated per period of rotation is
A
$${{{{\left( {B\pi r\omega } \right)}^2}} \over {2R}}$$
B
$${{{{\left( {B\pi {r^2}\omega } \right)}^2}} \over {8R}}$$
C
$${{B\pi {r^2}\omega } \over {2R}}$$
D
$${{{{\left( {B\pi r{\omega ^2}} \right)}^2}} \over {8R}}$$

Explanation

$$\phi = \overrightarrow B .\overrightarrow A ;\phi = BA\cos \,\omega t$$

$$\varepsilon = - {{d\phi } \over {dt}} = \omega BA\sin \,\omega t;\,\,$$

$$i = {{\omega BA} \over R}\sin \,\omega t$$

$${P_{inst}} = {t^2}R = {\left( {{{\omega BA} \over R}} \right)^2} \times R{\sin ^2}\omega t$$

$${p_{avg}} = {{\int\limits_0^T {{P_{inst}} \times dt} } \over {\int\limits_0^T {dt} }}$$

$$ = {{{{\left( {\omega BA} \right)}^2}} \over R}{{\int\limits_0^T {{{\sin }^2}\omega tdt} } \over {\int\limits_0^T {dt} }}$$

$$ = {1 \over 2}{{{{\left( {\omega BA} \right)}^2}} \over R}$$

$$\therefore$$ $${P_{abg}} = {{{{\left( {\omega B\pi {r^2}} \right)}^2}} \over {8R}}\,\,\,\,\,\left[ {A = {{\pi {r^2}} \over 2}} \right]$$
2

AIEEE 2004

MCQ (Single Correct Answer)
A coil having $$n$$ turns and resistance $$R\Omega $$ is connected with a galvanometer of resistance $$4R\Omega .$$ This combination is moved in time $$t$$ seconds from a magnetic field $${W_1}$$ weber to $${W_2}$$ weber. The induced current in the circuit is
A
$${{\left( {{W_2} - {W_1}} \right)} \over {Rnt}}$$
B
$$ - {{n\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rt}}$$
C
$$ - {{\left( {{W_2} - {W_1}} \right)} \over {5\,\,Rnt}}$$
D
$$ - {{n\left( {{W_2} - {W_1}} \right)} \over {Rt}}$$

Explanation

$${{\Delta \phi } \over {\Delta t}} = {{\left( {{W_2} - {W_1}} \right)} \over t}$$

$${R_{tot}} = \left( {R + 4R} \right)\Omega = 5R\Omega $$

$$i = {{nd\phi } \over {{R_{tot}}dt}} = {{ - n\left( {{W_2} - {W_1}} \right)} \over {5Rt}}$$

( as $${W_2}\,\,\& \,\,{W_1}$$ are magnetic flux )
3

AIEEE 2004

MCQ (Single Correct Answer)
In an $$LCR$$ series $$a.c.$$ circuit, the voltage across each of the components, $$L,C$$ and $$R$$ is $$50V$$. The voltage across the $$L.C$$ combination will be
A
$$100V$$
B
$$50\sqrt 2 $$
C
$$50$$ $$V$$
D
$$0$$ $$V$$ (zero)

Explanation

Since the phase difference between $$L$$ & $$C$$ is $$\pi ,$$

$$\therefore$$ net voltage difference across $$LC=50-50=0$$
4

AIEEE 2004

MCQ (Single Correct Answer)
Alternating current can not be measured by $$D.C.$$ ammeter because
A
Average value of current for complete cycle is zero
B
$$A.C.$$ Changes direction
C
$$A.C.$$ can not pass through $$D.C.$$ Ammeter
D
$$D.C.$$ Ammeter will get damaged.

Explanation

$$D.C.$$ ammeter measure average current in $$AC$$ current, average current is zero for complete cycle. Hence reading will be zero.

Questions Asked from Alternating Current and Electromagnetic Induction

On those following papers in MCQ (Single Correct Answer)
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