Electromagnetic Induction Test

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Electromagnetic Induction Test - 65

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Question 1
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A square wire frame of side $$a$$ is placed a distance $$b$$ away from a long straight conductor carrying current $$I$$. The frame has resistance $$R$$ and self inductance $$L$$. The frame is rotated by $$180^o$$ about OO' as shown in figure. Find the electric charge flown through the frame.

A
$$\displaystyle \dfrac{2 \mu_0 i a^2}{2 \pi Rb}$$

B
$$\displaystyle \dfrac{\mu_0 i}{2 \pi R} \log_e \dfrac{b + a}{b - a}$$

C
$$\displaystyle \dfrac{\mu_0 ia}{2 \pi R} \log_e \dfrac{b + a}{b - a}$$

D
none of these

Solution

$$\displaystyle i = \dfrac{1}{R} \left [ \dfrac{d \psi}{dt} + L \dfrac{di}{dt} \right ]$$
$$\displaystyle q = \int idt = \dfrac{1}{R} [\Delta \psi + 0] = \dfrac{\Delta \psi}{R} = \dfrac{1}{R} \int_{b - a}^{b + a} Ba.dx$$
$$\displaystyle = \dfrac{1}{R} \int_{b-a}^{b +a } \dfrac{\mu_0 ia}{2 \pi x} dx = \dfrac{\mu_0 ia}{2 \pi R} \log_e \dfrac{b + a}{b -a}$$
Question 2
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Find the inductance of a solenoid of length $$l_0$$, made of Cu windings of mass $$m$$. The winding resistance is equal to $$R$$. The diameter of solenoid << $$l$$. $$\rho_0$$ is resistivity of Cu and $$\rho$$ is density of the Cu.

A
$$\displaystyle \dfrac{\mu_0 Rm}{2 \pi l_0 \rho \rho_0}$$

B
$$\displaystyle \dfrac{\mu_0 Rm}{4 \pi l_0 \rho \rho_0}$$

C
$$\displaystyle \dfrac{\mu_0 Rm}{3 \pi l_0 \rho \rho_0}$$

D
$$\displaystyle \dfrac{2 \mu_0 Rm}{3 \pi l_0 \rho \rho_0}$$

Solution

$$R=\cfrac{\rho_0 l}{A}$$
Length of the wire $$l = \displaystyle \dfrac{RA}{\rho_0} = \dfrac{Rm}{\rho_0 \rho l} $$
or

$$\displaystyle l = \sqrt{\dfrac{Rm}{\rho \rho_0}}$$ where $$\rho_0$$ is

resistivity of the Cu and $$\rho$$ is density of the $$Cu l = nl_0 2

\pi r$$ and $$L = \mu_0 n^2 l_0 \pi r^2$$ where $$l_0$$ is length of

the solenoid and r is the radius of the solenoid. Then $$l = l_0 2 \pi r

\displaystyle \sqrt{\dfrac{L}{\mu_0 l_0 \pi r^2}}$$

or $$\displaystyle l = \sqrt{\dfrac{4 \pi L l_0}{\mu_0}}$$

or $$\displaystyle l = \sqrt{\dfrac{Rm}{\rho \rho_0}}$$

or $$\displaystyle L = \dfrac{\pi_0 Rm}{4 \pi l_0 \rho \rho_0}$$
Question 3
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Consider the situation shown. The wire AB is sliding on fixed rails with a constant velocity. If the wire AB is replaced by semi-circular wire, the magnitude of induced e.m.f will

A
increase

B
decrease

C
remain the same

D
increase or decrease depending on whether the semi-circle buldges towards the resistance or away from it

Solution

Faraday's Law

$$E = - \cfrac{d \phi}{dt}$$

$$ \phi = BA$$

Case I

$$ A_1 = Lx $$

Case 2

$$ A_2 = Lx + \pi r^2 /2$$

$$ E = - \cfrac{d \phi }{dt} = Blv$$

Same for both, since the rate of change of area is same for both the cases.
Question 4
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A metallic rod of length $$l$$ is hinged at the point $$M$$ and is rotating about an axis perpendicular to the plane of paper with a constant angular velocity $$\omega$$. A uniform magnetic field of intensity $$B$$ is acting in the region (as shown in the figure) parallel to the plane of paper. The potential difference between the points $$M$$ and $$N$$.

A
is always zero

B
varies between $$\cfrac { 1 }{ 2 } B\omega { l }^{ 2 }$$ to $$0$$

C
is always $$\cfrac { 1 }{ 2 } B\omega { l }^{ 2 }$$

D
is always $$ B\omega { l }^{ 2 }$$

Solution

Let an elementary part of width $$dx$$.
Total induced Emf across the rod is given by $$E = \int_0^l (\vec{v} \times \vec{B}) . \vec{dx} $$
As according to the figure, $$\vec{B} = B \hat{x}$$ and $$\vec{v} = -v \hat{y} = -wx \hat{y}$$
$$\therefore$$ $$\vec{v} \times \vec{B} = wxB \hat{k}$$
$$\implies (\vec{v} \times \vec{B}) . \vec{dx} = (wxB \hat{k}) . {dx}$$ $$ \hat{x} = 0$$
Note: At every instant, $$(\vec{v} \times \vec{B}) \perp $$ $$\vec{dx}$$
Thus total Emf induced is always $$Zero.$$
Question 5
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A conducting rod of length $$l$$ falls vertically under gravity in a region of uniform magnetic field $$\vec { B } $$. The field vectors are inclined at an angle $$\theta$$ with the horizontal as shown in figure. If the instantaneous velocity of the rod is $$v$$, the induced emf in the rod $$ab$$ is:

A
$$Blv$$

B
$$Blv\cos { \theta } $$

C
$$Blv\sin { \theta } $$

D
zero

Solution

$$E = \vec{B}.(\vec{v} \times l)$$
$$E = Bvl sin \theta $$
if v is parallel to l
then
$$E =0$$
Question 6
1 / -0

The potential difference across a $$150mH$$ inductor as a function of time is shown in figure. Assume that the initial value of the current in the inductor is zero. What is the current when $$t=4.0ms$$?

A
$$2.67 \times 10^{-4}$$A

B
$$3.67 \times 10^{-2}$$A

C
$$6.67 \times 10^{-2}$$A

D
$$9.67 \times 10^{-4}$$A

Solution

$${V}_{L}=L\cfrac { di }{ dt } $$

$$\therefore \quad di=\cfrac { 1 }{ L } \left( { V }_{ L }dt \right) $$

$$\therefore \quad \int { di=i= } \cfrac { 1 }{ L } \int { { V }_{ L }dt } \quad or\quad i=\cfrac { 1 }{ L } $$

At $$t=4ms$$

$$i={ \left( 150\times { 10 }^{ -3 } \right) }^{ -1 }\left( \cfrac { 1 }{ 2 } \times 4\times { 10 }^{ -3 }\times 5 \right) =6.67\times { 10 }^{ -2 }A\quad $$
Question 7
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An alternating current $$I$$ in an inductance coil varies with time $$t$$ according to the graph as shown:
Which one of the following graphs gives the variation of voltage with time?

A

B

C

D

Solution

emf = $$ L \dfrac{di}{dt} $$
Rate of change of current is constant for one period at a positive value and is constant at negative value for the second time period.
Therefore emf is a constant positive value for first half and constant negative value for second half. Option C is correct.
Question 8
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A loop of area $$1{m}^{2}$$ is placed in a magnetic field $$B=2T$$, such that plane of the loop is parallel to the magnetic field. If the loop is rotated by $${180}^{o}$$, the amount of net charge passing through any point of loop, if its resistance is $$10\Omega$$ is :

A
$$0.4C$$

B
$$0.2C$$

C
$$0.8C$$

D
$$0C$$

Solution

$${ \phi }_{ i }=BS\cos { { 0 }^{ o } } =2Wb$$
$${ \phi }_{ f }=BS\cos { { 180 }^{ o } } =-2Wb$$
$$\left| \Delta \phi \right| =4Wb$$
$$\left| \Delta q \right| =\cfrac { \left| \Delta \phi \right| }{ R } =\cfrac { 4 }{ 10 } =0.4C$$
Question 9
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A semi-circular conducting ring $$acb$$ of radius $$R$$ moves with constant speed $$v$$ in a plane perpendicular to uniform magnetic field $$B$$ as shown in figure. Identify the correct statement

A
$${V}_{a}-{V}_{c}=BRv$$

B
$${V}_{b}-{V}_{c}=BRv$$

C
$${V}_{a}-{V}_{b}=0$$

D
None of these

Solution

$$E = _0 ^{\theta}\int \vec{B}.(\vec{v} \times dl)$$

Consider a wire AB

$$E = _0 ^{180^o}\int \vec{B}.(\vec{v} \times r d \theta sin \theta)$$

$$ \vec{v}$$ is parallel to $$l$$
Hence $$V_A- V_B =0$$
Also

$$V_C- V_A = _0 ^{90^o}\int \vec{B}.(\vec{v} \times r d \theta sin \theta)$$

$$ V_C-V_A = BvR$$

$$V_C - V_B = BvR$$
$$V_A - V_B= 0$$
Question 10
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A straight conducting rod $$PQ$$ is executing SHM in $$xy$$ plane from $$x=-d$$ to $$x=+d$$. Its mean position is $$x=0$$ and its length is along y-axis. There exists a uniform magnetic field $$B$$ from $$x=-d$$ to $$x=0$$ pointing inward normal to the paper and from $$x=0$$ to $$=+d$$ there exists another uniform magnetic field of same magnitude $$B$$ but pointing outward normal to the plane of the paper. At the instant $$t=0$$, the rod is at $$x=0$$ and moving to the right. The induced emf $$(\varepsilon )$$ across the rod $$PQ$$ vs time $$(t)$$ graph will be

A

B

C

D

Solution

The displacement equation of the rod executing $$SHM$$ $$x = d $$ $$ sinwt$$ .......(1)
where $$d$$ is the amplitude of oscillation.

Thus velocity equation is given by $$v = \dfrac{dx}{dt} = wd$$ $$coswt$$

Let the length of the conduction rod $$PQ$$ be $$L$$

Induced EMF in the rod $$\mathcal{E} = BvL = BLwd$$ $$coswt$$

$$\implies$$ At $$t = 0$$, Induced EMF is maximum i.e

$$\mathcal{E_{max}} = BLwd$$ and after that , it decreases.

At $$t = \dfrac{T}{4}$$, Induced EMF $$\mathcal{E} = 0$$

Now the rod starts to move in $$-ve$$ x direction and reach the mean position i.e $$x= 0$$

At $$t \leq \dfrac{T}{2}$$, Induced EMF $$\mathcal{E} = -BLwd$$

Now for $$t> \dfrac{T}{2}$$ OR $$x<0$$, $$B$$ reverses its direction and hence $$\mathcal{E}$$ also reverses its polarity.

$$\implies $$At $$t > \dfrac{T}{2}$$, Induced EMF $$\mathcal{E} = BLwd$$ and then it decreases with time again.