Electromagnetic Induction Test

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Electromagnetic Induction Test - 69

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Question 1
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An electric bulb has a rated power of $$50 W$$ at $$100 V$$. If it is used on an AC source $$200 V$$, $$50 Hz$$, a choke has to be used in series with it. This choke should have an inductance of

A
$$0.01 mH$$

B
$$1 mH$$

C
$$0.1 H$$

D
$$1.1 H$$

Solution

Resistance of the bulb $$=\dfrac{V^2}{P}=200\Omega$$

Let inductance of choke be $$L$$.

Then the impendence of inductor $$=2\pi\times50\times L$$

Both the voltages will be perpendicular in the phasor diagram. Net voltage is vector addition of both.
The total is $$200V$$. For $$100V$$ to be on resistor, $$100\sqrt{3}$$ will be on inductor.

$$\Rightarrow \dfrac{2\pi\times50\times L}{200}=\sqrt{3}$$

$$\Rightarrow L=1.1H$$
Question 2
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A uniform magnetic field exists in region given by $$\vec{B}=3\hat{i}+4\hat{j}+5\hat{k}$$. A rod of length $$5$$m is placed along y-axis is moved along x-axis with constant speed $$1$$m$$/$$sec. Then induced e.m.f. in the rod will be.

A
Zero

B
$$25$$v

C
$$20$$v

D
$$15$$v

Solution

Given : $$\vec{B} = 3\hat{i}+4\hat{j}+5\hat{k}$$
Length of rod $$\vec{l}= 5\hat{j} $$
Velocity of rod $$\vec{v} = 1\hat{i} \ $$
Induced emf is given by $$\mathcal{E} = (\vec{v}\times \vec{B}).\vec{l}$$
$$\therefore$$ $$\mathcal{E} = (1\hat{i}\times (3\hat{i}+4\hat{j}+5\hat{k})].(5\hat{j})$$
Or $$\mathcal{E} = (4\hat{k}-5\hat{j}).(5\hat{j}) = -25$$ volts
$$\implies \ |\mathcal{E} | = 25$$ volts
Question 3
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A rod of length $$b$$ moves with a constant velocity $$v$$ in the magnetic field of a straight long conductor that carries a current $$i$$ as shown in the figure. The emf induced in the rod is

A
$$\dfrac {\mu_{0}iv}{2\pi}\tan^{-1}\dfrac {a}{b}$$

B
$$\dfrac {\mu_{0}iv}{2\pi} ln \left (1 + \dfrac {b}{a}\right )$$

C
$$\dfrac {\mu_{0}iv\sqrt {ab}}{4\pi (a + b)}$$

D
$$\dfrac {\mu_{0}iv(a + b)}{4\pi ab}$$

Solution
Question 4
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A coil of area 500 $$cm^2$$ having 1000 turns is placed such that the plane of the coil is perpendicular to a magnetic field of magnitude $$4 \times 10^{-5}$$ $$weber/m^2$$. If it is rotated by 180 about an axis passing through one of its diameter in 0.1 sec, find the average induced emf.

A
zero.

B
30 mV

C
40 mV

D
50 mV

Solution

0iven that :- $$N=1000, B=4\times 10^{-5}weber/m^2, A=500cm^2=0.05m^2$$

Initial flux linked with the coil, $$\phi_1=1000\times 4\times 10^{-5}\times 0.05$$

$$\implies \phi_1=2\times 10^{-3}weber$$

After rotation of $$180^{o}$$, B remains same but normal vector gets reversed, hence $$\phi_2=-\phi_1$$

Average EMF=$$E=\dfrac{-\Delta \phi}{t}$$

$$\implies E=-\dfrac{\phi_2-\phi_1}{t}$$

$$\implies E=\dfrac{\phi_1-\phi_2}{t}$$

$$\implies E=\dfrac{2\phi_1}{t}$$

$$\implies E=\dfrac{4\times 10^{-3}}{0.1}V$$

$$\implies E=40mV$$

Answer-(C)
Question 5
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Two coils have mutual inductance $$0.005 H$$. The current changes in the form coil according to equation, $$ I = I_0 \sin \omega t . $$ Where $$ I_0 = 10 A. $$ and $$ \omega = 100 \pi $$ rads/s. The maximum value of emf in the second coil is :

A
$$ 12 \pi $$

B
$$ 8 \pi $$

C
$$ 5 \pi $$

D
$$ 2 \pi $$

Solution

Mutual inductance between two coils
M = 0.005 H
Peak current $$ l_0 = 10 A $$
Angular frequency $$ \omega = 100 \pi $$ rad/s
Current $$ l = l_0 \sin \omega t $$
$$ \dfrac {d}{dt} = \dfrac {d}{dt} ( l \sin \omega t ) $$
$$ = l_0 \cos \omega t . \omega $$
$$ = 10 \times 1 \times 100 \pi $$
$$ = 1000 \pi $$
Hence, induced emf is given by
$$ E = M \times \dfrac {dl}{dt} $$
$$ = 0.005 \times 1000 \times \pi = 5 \pi V $$
Question 6
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A coil of area $$500 cm^2$$ having 1000 turns is placed such that the plane of the coil is perpendicular to a magnetic field of magnitude $$4 10^{-5} weber/m^2$$. If it is rotated by $$180$$ degree about an axis passing through one of its diameter in $$0.1$$ sec, find the average induced emf.

A
$$zero.$$

B
$$30 mV$$

C
$$40 mV$$

D
$$50 mV$$

Solution

Area of the coil $$A = 500 \ cm^2 = 0.05 \ m^2$$
Number of coil $$N = 1000$$
Magnetic field $$B = 4\times 10^{-5}   \ Wb/m^2$$
Magnetic flux passing through the coil initially $$(\theta = 0^o)$$,
$$\phi_i = NBA\cos\phi =1000\times 4\times 10^{-5}\times 0.05\times \cos 0^o =0.002 \ Wb $$
Magnetic flux passing through the coil finally $$(\theta = 180^o)$$,
$$\phi_f = NBA\cos\phi =1000\times 4\times 10^{-5}\times 0.05\times \cos 180^o =-0.002 \ Wb $$
Thus change of flux $$\Delta \phi =\phi_i-\phi_f = 0.002-(-0.002) = 0.004 \ Wb$$
Change in time $$\Delta t = 0.1 \ s$$
Thus average emf induced $$\mathcal{E} = \dfrac{\Delta \phi}{\Delta t} = \dfrac{0.004}{0.1} = 40 \ mV$$
Question 7
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Radii of two conducting circular loops are $$b$$ and $$a$$ respectively, where $$b > > a$$. Centres of both loops coincide but planes of both loops are perpendicular to each other. The value of mutual inductance for these loops.

A
$$\dfrac {\mu_{0}\pi b^{2}}{2a}$$

B
zero

C
$$\dfrac {\mu_{0}\pi ab}{2(a + b)}$$

D
$$\dfrac {\mu_{0}\pi a^{2}}{2b}$$

Solution

At the position of smaller loop, magnetic field due to larger loop is parallel to the plane of smaller loop. Due to larger loop, magnetic flux with smaller loop is zero. Hence, mutual induction is zero.
$$\phi = M\ i_{2}$$
$$\Rightarrow 0 = Mi_{2}$$
$$\Rightarrow M = 0$$.
Question 8
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A generator with a circular coil of $$100$$ turns of area $$2\times { 10 }^{ -2 }{ m }^{ 2 }$$ is immersed in a $$0.01T$$ magnetic field and rotated at a frequency of $$50Hz$$. The maximum emf which is produced during a cycle is

A
$$6.28V$$

B
$$3.44V$$

C
$$10V$$

D
$$1.32V$$

Solution

Given : $$N = 100$$ turns $$A = 2\times 10^{-2}m^2$$ $$B = 0.01 \ T$$ $$\nu = 50 \ Hz$$
So $$w =2\pi \nu = 2\times \dfrac{22}{7}\times 50$$
Maximum emf induced
$${ e }_{ 0 }=NBA\omega $$
$$e_0=100\times 0.01\times 2\times { 10 }^{ -2 }\times 2\times \cfrac { 22 }{ 7 } \times 50=6.28V$$
Question 9
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A wire as a parabola $$y = 4x^2$$ is located in a uniform magnetic field of inductance B perpendicular to the XY plane. At t=0 a connection starts translation wise from the parabola apex with constant acceleration $$\alpha$$. The induced emf in the loop, thus formed, as a function of y is:

A
$$e = \sqrt{2\alpha}\cdot{By}$$

B
$$e = By\sqrt\frac{\alpha}{2}$$

C
$$e = \frac{By\sqrt\alpha}{2\sqrt2}$$

D
$$e = \frac{ By\sqrt\alpha}{4}$$

Solution

$$ \overrightarrow { B } =B\uparrow $$
Velocity of rod after travelling $$dy$$ distance $$u$$.
we know that,
$$S=ut+\dfrac { 1 }{ 2 } a{ t }^{ 2 }$$
$$ y=0+\dfrac { 1 }{ 2 } \alpha { t }^{ 2 }$$ ($$t=\sqrt{\dfrac{2y}{\alpha}}$$)
$$ \dfrac { dy }{ dt } =\alpha t $$
$$ V=\alpha t=\sqrt { 2\alpha y } $$
emf induced $$ \varepsilon =VB(2x)$$
$$=\sqrt { 2\alpha y } \times 2xB $$
$$ x=\dfrac { \sqrt { y } }{ 2 } $$
$$ \varepsilon =\sqrt { 2\alpha } .By$$
Question 10
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A superconducting rigid planar loop of area $$A$$ and self-inductance $$L$$ carrying a current is held motionless in a region of free spaces. Now a uniform magnetic field of induction $$B$$-pointing everywhere parallel to the magnetic moment $$m$$ of the loop is switched on. Current in the loop after the magnetic field is switched on us given by

A
$$\dfrac {AB}{L}$$

B
$$\dfrac {m}{A}$$

C
$$\dfrac {m}{A}-\dfrac {AB}{L}$$

D
$$\dfrac {m}{A}+\dfrac {AB}{L}$$

Solution