Electromagnetic Induction Test

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Electromagnetic Induction Test - 67

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Question 1
1 / -0

Magnetic flux linked with a stationary loop resistance $$R$$ varies with respect to time during the time period $$T$$ as follows:
$$\phi=at(T-t)$$
The amount of heat generated in the loop during that time (inductance of the coil is negligible) is

A
$$\cfrac{\alpha T}{3R}$$

B
$$\cfrac{{a}^{2}{T}^{2}}{3R}$$

C
$$\cfrac{{a}^{2}{T}^{2}}{R}$$

D
$$\cfrac{{a}^{2}{T}^{3}}{3R}$$

Solution

Given , $$\phi=at(T-t)$$

The emf is, $$\varepsilon=-\dfrac{d\phi}{dt}=-at+2at=at$$

Power, $$P=\varepsilon I=\varepsilon . \dfrac{\varepsilon}{R}=\dfrac{a^2t^2}{R}$$

Heat generated, $$H=\int_0^T Pdt=\int^T_0\dfrac{a^2t^2}{R} dt=\dfrac{a^2T^3}{3R}$$
Question 2
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In the figure, there exists a uniform magnetic field $$B$$ into the plane of paper. Wire $$CD$$ is in the shape of an arc and is fixed. $$OA$$ and $$OB$$ are the wires rotating with angular velocity $$\omega$$ as shown in the figure in the same plane as that of the arc about point $$O$$. If at some instant, $$OA=OB=1$$ and each wire makes angle $$\theta={30}^{o}$$ with y-axis, then the current through resistance $$R$$ is (wire $$OA$$ and $$OB$$ have no resistance)

A
$$0$$

B
$$\cfrac{B\omega {l}^{2}}{R}$$

C
$$\cfrac{B\omega {l}^{2}}{2R}$$

D
$$\cfrac{B\omega {l}^{2}}{4R}$$

E
answer required

Solution

EMF induced in wire OA = $$ \dfrac{1}{2} B l^2 \omega $$ with A positive wrt O

emf induced in wire OB = $$ \dfrac{1}{2} B l^2 \omega $$ with O positive wrt B

Therefore total emf in the loop consisting of wire OA , OB and the arc is $$ B l^2 \omega $$

current in loop = $$ \dfrac{ B l^2 \omega}{R}$$
Question 3
1 / -0

The current through the coil in figure (i) varies as shown in figure (ii). Which graph best shows ammeter $$A$$ reading as a function of time?

A

B

C

D

E
answer required

Solution

When current flows through first circuit, flux links the second circuit due to mutual inductance. when there is a change in current, there is a corresponding changing in linking flux between the two circuits and hence a emf is induced in the second circuit given by,
emf = rate of change of flux = M $$ \times $$ rate of change of current.
current in the first circuit changes twice with a constant rate, one time positive rate of change and second time negative rate of change.
Therefore emf will be constant positive for first period when current changes. And emf will be constant negative in the second period the current changes.
Question 4
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A conducting ring of radius $$r$$ is rolling without slipping with a constant angular velocity $$\omega$$ (figure). If the magnetic field strength is $$B$$ and is directed into the page then the emf induced across $$PQ$$ is

A
$$B\omega {r}^{2}$$

B
$$\cfrac{B\omega{r}^{2}}{2}$$

C
$$4B\omega{r}^{2}$$

D
$$\cfrac{B\omega{r}^{2}}{4}$$

Solution

Let the velocity of each point due to rotation form an angle $$\theta$$ with the vertical.
EMF induced across PQ $$=\int_{0}^{\pi/2} B(r\omega\cos\theta)(rd\theta)$$
$$=\int_0^{\pi/2}Br^2\omega\cos\theta d\theta\\ =B\omega r^2$$
Question 5
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A rectangular loop with a sliding conductor of length $$l$$ is located in a uniform magnetic field perpendicular to the plane of the loop (figure). The magnetic induction is $$B$$. The conductor has a resistance $$R$$. The sides $$AB$$ and $$CD$$ have resistances $${R}_{1}$$ and $${R}_{2}$$ respectively. Find the current through the conductor during its motion to the right with a constant velocity $$v$$.

A
$$\cfrac { Blv({ R }_{ 1 }+{ R }_{ 2 }) }{ { R }_{ 1 }({ R }_{ 1 }+{ R }_{ 2 }) } $$

B
$$\cfrac { B{ l }^{ 2 }v }{ { R }_{ 1 }+{ R }_{ 1 }{ R }_{ 2 } } $$

C
$$\cfrac { B{ l }^{ }v({ R }_{ 1 }+{ R }_{ 2 }) }{ { R }_{ 1 }{ R }_{ 2 }+R({ R }_{ 1 }+{ R }_{ 2 }) } $$

D
$$\cfrac { B{ l }^{ 2 }v({ R }_{ 1 }+{ R }_{ 2 }) }{ { R }_{ 1 }{ R }_{ 2 }+R({ R }_{ 1 }+{ R }_{ 2 }) } $$

Solution

$${ R }_{ 1 }\quad and\quad { R }_{ 2 }\quad are\quad in\quad parallel\quad the\quad this\quad combination\quad is\quad in\quad series\quad with\quad R.\\ $$

$$Thus\quad Resistance\quad =\quad \dfrac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } +R\\$$

$$ EMF=Blv\\$$
$$ i\quad =\quad \dfrac { Blv }{ \dfrac { { R }_{ 1 }{ R }_{ 2 } }{ { R }_{ 1 }+{ R }_{ 2 } } +R } =\dfrac { Blv({ R }_{ 1 }+{ R }_{ 2 }) }{ R({ R }_{ 1 }+{ R }_{ 2 })+{ R }_{ 1 }{ R }_{ 2 } } $$
Question 6
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A rectangular frame $$ABCD$$ made of a uniform metal wire has a straight connection between $$E$$ and $$F$$ made of the same wire as shown in figure. $$AEFD$$ is a square of side $$1m$$ and $$EB=FC=0.5m$$. The entire circuit is placed in a steadily increasing, uniform magnetic field directed into the plane of paper and normal to it. The rate of change of the magnetic field is $$1$$ $$T{s}^{-1}$$. The resistance per unit length of the wire is $$1\Omega$$ $${m}^{-1}$$. Find the magnitude and direction of the current in the segment $$AE$$, $$BE$$ and $$EF$$.

A
$$ \dfrac{6}{22}$$A, E to A; $$\dfrac{7}{22}$$A, B to E; $$\dfrac{1}{22}$$A, F to E

B
$$ \dfrac{1}{22}$$A, E to A; $$\dfrac{6}{22}$$A, E to B; $$ \dfrac{7}{22}$$A, F to E

C
$$\dfrac{7}{22} $$A, E to A; $$\dfrac{6}{22}$$A, B to E; $$\dfrac{1}{22}$$A, F to E

D
zero, zero, zero

Solution

flux through $$AEFD, \, \phi_1 = BA$$ & through $$EBCF$$ $$\phi_2 = BA_2 $$
$$A_1 = 1\times 1 = 1m^2$$
& $$A_2 = 1\times \dfrac{1}{2}m^2 = \dfrac{1}{2}m^2$$
Also, the magnitude of induced emf will be
$$e_1 = \left|\dfrac{d\phi_1}{dt}\right| = A_1 \dfrac{dB}{dt} = 1V$$ (given $$\dfrac{dB}{dt} = 1$$)

$$e_2 = \left| \dfrac{d\phi_2}{dt}\right| = A_2 = \dfrac{dB}{dt} = \dfrac{1}{2} $$ or $$0.5V$$

Since, the magnetic fd. is increasing & directed into the paper, hence the induced current in these loops should be in anti-clockwise dirt.
The equivalent circuit will be given as
On applying kirchoff's voltage.
Law in these loops
$$4i_1 - i_2 = e_1$$
$$3i_2 - i_1 = e_2$$

$$i_1 = \dfrac{7}{22} A$$
$$i_2 = \dfrac{6}{22}A$$
$$i_1 - i_2 = \dfrac{1}{22}A$$

$$i_{AE} = i_1,\, i_{BE} = i_2, \, i_{EF} = i_1 - i_2$$
Question 7
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In the circuit shown in figure, a conducting wire $$HE$$ is moved with a constant speed $$v$$ toward left. The complete circuit is placed in a uniform magnetic field $$\vec { B } $$ perpendicular to the plane of the circuit inward. The current in $$HKDE$$ is

A
clockwise

B
anticloclwise

C
alternating

D
zero

E
answer required

Solution

Assuming the $$AK$$ and $$BD$$ are infinitely long
By Faraday's Law
$$E = \cfrac{d\phi}{dt} = BvL$$
$$E_{HE}= BvL$$
The analogous diagram is shown in the figure below
Resistor and capacitor are in parallel
$$ E - V_c = 0$$
$$V_C = Bvl$$
and
$$Q = CV = CE =const$$
$$i_C=\cfrac{dQ}{dt}=0$$
Question 8
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A conducting rod $$PQ$$ of length $$l=2m$$ is moving at a speed of $$2m$$ $${s}^{-1}$$ making an angle of $${30}^{o}$$ with its length. A uniform magnetic field $$B=2T$$ exists in a direction perpendicular to the plane of motion. Then

A
$${V}_{P}-{V}_{Q}=8V$$

B
$${V}_{P}-{V}_{Q}=4V$$

C
$${V}_{Q}-{V}_{P}=8V$$

D
$${V}_{Q}-{V}_{P}=4V$$

Solution

$$E = BvLsin(\theta)$$
$$E = 2\times 2 \times 2 sin(30^o) = 4 V$$
By right hand rule:
$$V_P> V_Q$$
$$V_P -V_Q = 4 V$$
Question 9
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Two identical conducting rings $$A$$ and $$B$$ of radius $$R$$ are rolling over a horizontal conducting plane with same speed $$v$$ but in opposite direction. A constant magnetic field $$B$$ is present pointing into the plane of the paper. Then the potential difference between the highest points of the two rings is

A
$$0$$

B
$$2BvR$$

C
$$4BvR$$

D
none of these

E
answer required

Solution

Let us consider one ring, travelling in one direction. Each ring has two components of velocity, the linear component and the angular component.
The potential difference induced between the top and bottom points of each ring due to the angular component is zero.
The potential difference induced between the same points due to linear component $$=\int_{0\\-\pi/2}^{\pi/2} B(vcos\theta)Rd\theta$$
$$=2BvR$$
This is the potential difference between top and bottom points of one ring.
Similarly the potential difference between top and bottom of other ring is $$2BvR$$, however the polarity is opposite since the rings travel in opposite directions. The potential of the bottom of both ring is equal(connected).
Hence potential difference between top of both the rings $$=2BvR-(-2BvR)=4BvR$$
Question 10
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A rod of length $$L$$ rotates in the form of a conical pendulum with an angular velocity $$\omega$$ about its axis as shown in figure. The rod makes an angle $$\theta$$ with the axis. The magnitude of the motional emf developed across the two ends of the rod is

A
$$\cfrac{1}{2}B\omega {L}^{2}$$

B
$$\cfrac{1}{2}B\omega {L}^{2}\tan ^{ 2 }{ \theta } $$

C
$$\cfrac{1}{2}B\omega {L}^{2}\cos ^{ 2 }{ \theta } $$

D
$$\cfrac{1}{2}B\omega {L}^{2}\sin ^{ 2 }{ \theta } $$

E
answer required

Solution

Figure shows the Top view, in which the rod of length $$r$$ is revolving in a circle about an axis passing through its one end.
From top view, effective length $$=r=l\sin { \theta } $$
Induced emf $$ e=\dfrac { 1 }{ 2 } B\omega { r }^{ 2 }=\dfrac { 1 }{ 2 } B\omega { l }^{ 2 }\sin ^{ 2 }{ \theta } $$