Electromagnetic Induction Test

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Electromagnetic Induction Test - 17

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Question 1
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Which of the following units denotes the dimensions $${M}{L}^{2}/{Q}^{2}$$, where $${Q}$$ denotes the electric charge?

A
$$Weber$$ ($$Wb$$)

B
$$Wb$$$$/\mathrm{m}^{2}$$

C
$$Henry$$ ($$H$$)

D
$$\mathrm{H}/\mathrm{m}^{2}$$

Solution

Weber $$=ML^2T^{-2}I^{-1}$$
$$=ML^2T^{-2}Q^{-1}T=ML^2T^{-1}Q^{-1}$$ ($$I=QT^{-1}$$)
Henry H is SI unit of inductance.
$$H=ML^2T^{-2}I^{-2}$$ also $$I=QT^{-1}$$
so $$H=ML^2T^{-2}Q^{-2}T^2=ML^2Q^{-2}$$
Question 2
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Consider a thin metallic sheet perpendicular to the plane of the paper moving with speed $$'v'$$ in a uniform magnetic field B going into the plane of the paper (See figure). If charge densities $$\sigma_1$$ and $$\sigma_2$$ are induced on the left and right surfaces, respectively, of the sheet then (ignore fringe effects) :

A
$$\sigma_1=\frac{-\epsilon_0vB}{2},\ \sigma_2=\frac{\epsilon_0vB}{2}$$

B
$$\sigma_1=\epsilon_0vB,\ \sigma_2 = -\epsilon_0vB$$

C
$$\sigma_1=\frac{\epsilon_0vB}{2},\ \sigma_2=\frac{-\epsilon_0vB}{2}$$

D
$$\sigma_1=\sigma_2=\epsilon_0vB$$

Solution
Question 3
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A planar loops of wire rotates in a uniform magnetic field. Initially, at $$t = 0$$, the plane of the loop is perpendicular to the magnetic field. If it rotates with a period of $$10 s$$ about an axis in its plane then the magnitude of induced emf will be maximum and minimum, respectively at :

A
$$5.0 s$$ and $$7.5 s$$

B
$$2.5 s$$ and $$7.5 s$$

C
$$2.5 s$$ and $$5.0 s$$

D
$$5.0 s$$ and $$10.0 s$$

Solution

Time period, $$T = 10 s$$
Ref. image I

Axis of rotation $$\rightarrow $$ y-axis.

$$\vec{B} = B(-\hat{k})$$

ref. image II
at some time t, area vector makes angle '$$\theta$$' with $$\vec{B}$$.
flux, $$\phi = \vec{B}. \vec{A}$$

$$= BA \cos \theta$$

emf, $$\varepsilon= \left|\dfrac{-d \phi}{dt} \right| = \left(BA \sin \theta \dfrac{d \theta}{dt}\right)$$

$$= BA \omega \sin \theta$$

When $$\theta = \dfrac{\pi}{2} , \varepsilon = \varepsilon_{max}$$

So, $$\theta = n \pi + \dfrac{\pi}{2} $$ n = integes

When $$\theta = 0, \varepsilon = \varepsilon_{min} = 0$$

So, $$\theta = n \pi$$, n = integes

for $$\varepsilon = \varepsilon_{max}, \theta = \omega t = n \pi + \dfrac{\pi}{2}$$

$$\Rightarrow t = \dfrac{n \pi}{\omega} + \dfrac{\pi}{2 \omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} + \dfrac{\pi}{2 \times \left(\dfrac{2 \pi}{T} \right)}$$

$$t = \left(\dfrac{n \pi}{\dfrac{2 \pi}{T}} \right) + \dfrac{\pi}{2 \left(\dfrac{2 \pi}{T} \right)}$$

$$= \dfrac{n T}{2} + \dfrac{T}{4} = \dfrac{n \times 10}{2} + \dfrac{10}{4}$$
$$= 5n + 2.5$$

$$= 2.5 s, 5 + 2.5 , 5 \times 2 + 2.5 ...$$

$$= 2.5 s, 7.5 s, 12.5 s....$$

for $$\varepsilon = \varepsilon_{min}, \theta = \omega t = n \pi$$

$$\Rightarrow t = \dfrac{n \pi}{\omega} = \dfrac{n \pi}{\left(\dfrac{2 \pi}{T} \right)} = \dfrac{n T}{2}$$

$$\Rightarrow t = \dfrac{n \times 10}{2} = 5n$$

$$= 0 s, 5s, 10s, ...$$

So, possible option is option (C).
Question 4
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When the current changes from $$+2 A$$ to $$- 2 A$$ in $$0.05$$ second, an e.m.f. of $$8 V$$ is induced in a coil. The coefficient of self-induction of the coil is :

A
$$0.2 H$$

B
$$0.4 H$$

C
$$0.8 H$$

D
$$0.1 H$$

Solution

lf $$\mathrm{e}$$ is the induced $$\mathrm{e}$$.m.f. in the coil, then $$\displaystyle \mathrm{e}=-\mathrm{L}\frac{\mathrm{d}\mathrm{i}}{\mathrm{d}\mathrm{t}}$$
Therefore, $$\displaystyle \mathrm{L}=-\frac{\mathrm{e}}{\mathrm{d}\mathrm{i}/\mathrm{d}\mathrm{t}}$$
Substituting values, we get $$\displaystyle \mathrm{L}=\frac{-8\times 0.05}{-4}=0.1\mathrm{H}$$
Question 5
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A boat is moving due east in a region where the earth's magnetic field is $$5.0 \times 10^{-5}{N}{A}^{-1}{m}^{-1}$$ due north and horizontal. The boat 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
$$1$$ $$\mathrm{m}\mathrm{V}$$

B
$$0.75$$ $$\mathrm{m}\mathrm{V}$$

C
$$0.50$$ $$\mathrm{m}\mathrm{V}$$

D
$$0.15$$ $$\mathrm{m}\mathrm{V}$$

Solution

Induced emf is given by:
$$\varepsilon=$$ $$Bvl$$
On putting the values we get
$$=5\times 10^{-5}\times 1.50\times 2$$
$$= 0.15 \mathrm{m}\mathrm{V}$$
Question 6
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When current is coil changes from 5 A to 2 A in 0.1 s, an average of 50 V is produced. The self-inductance of the coil is.

A
6 H

B
0.67 H

C
1.67 H

D
3 H

Solution

Inudctor opposes change in curren, as the current is decreased, the inductor will convert magnetic energy into electrical by creating a potential difference. The voltage drop across inductor is given by

$$V=L\dfrac{di}{dt}$$

$$ -50=L \dfrac{2-5}{.1}$$ here negative sign is because voltage is increasing instead of drop across inductor

$$ -50=L \dfrac{2-5}{.1}$$

$$L=1.67 \;H $$.
Question 7
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At time $$t = 0$$ magnetic field of $$1000$$ Gauss is passing perpendicularly through the area defined by the closed loop shown in the figure. If the magnetic field reduces linearly to $$500$$ Gauss, in nest $$5s$$, then induced EMF in the loop is:

A
$$36\mu V$$

B
$$28\mu V$$

C
$$56\mu V$$

D
$$48\mu V$$

Solution

Area of loop $$AEBCFDA =$$ area of rectangle $$ABCD \ -$$ Area of triangle $$ABE\ -$$ Area of triangle $$DCF$$

$$\therefore$$ Required area $$=16cm \times 4 cm - \dfrac{1}{2}\times 4cm \times 2cm - \dfrac{1}{2}\times 4cm \times 2cm$$

$$= 64cm^2 - 4cm^2 - 4cm^2 = 56cm^2$$

Magnetic flux passing through an area is given by

$$\displaystyle \phi = \int B.dA = \int BdA \cos \theta$$.

Here, $$\theta = 0$$ and $$B$$ is constant throughout the area.
$$\therefore \phi_{initial} = B_{initial}.A = 1000 \, Gauss \times 56cm^2$$

$$= 56000 \, gauss\, cm^2$$

$$= 5.6\times 10^{-4}Wb$$

$$\begin{bmatrix} 1 \, Gauss = 10^{-4}T \, and \, 1cm^2 = 10^{-4}m^2\\ \therefore 1\, Gauss\, cm^2 = 10^{-8}Tm^2 = 10^{-8}Wb \end{bmatrix}$$

Also, $$\phi _{final} = B_{final}. A = 500\, Gauss \times 56cm^2$$

$$= 28000\, Gauss\, cm^2 = 2.8\times 10^{-4}Wb$$

Since the flux changes linearly,
$$\in = \left|\dfrac{d\phi}{dt}\right| = \dfrac{|\phi_{final} - \phi_{initial}|}{\Delta t} = \dfrac{(5.6-2.8)\times 10^{-4}}{5}V$$

$$= 56\times 10^{-6}V = 56\mu v$$
Question 8
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A long solenoid has $$1000\ $$turns. When a current of $$4\ A$$ flows through it, the magnetic flux linked with each turn of the solenoid is $$4\times 10^{-3}$$Wb. The self-inductance of the solenoid is:

A
$$4$$ H

B
$$3$$ H

C
$$2$$ H

D
$$1$$ H

Solution

Flux linked with each turn $$=4\times 10^{-3}\ Wb$$

$$\Rightarrow $$ Total flux linked $$=1000\times 4\times 10^{-3}\ Wb$$

$$\Rightarrow \phi_{total}=4$$

$$\Rightarrow LI=4$$

$$\Rightarrow L=1 \ H$$
Question 9
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A thin semicircular conducting the ring $$(PQR)$$ of radius $$'r'$$ is falling with its plane vertical in a horizontal magnetic field $$B$$, as shown in figure. The potential difference developed across the ring when its speed is $$v$$, is

A
$$Zero$$

B
$$Bv\pi r^2/2$$ and $$P$$ is at higher potential

C
$$\pi rBv$$ and $$R$$ is at higher potential

D
$$2rBv$$ and $$R$$ is at higher potential

Solution

$$emf= VBl_{eq} $$
$$= VB(2R)$$
where R is at higher potential and P is at lower potential
Question 10
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A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced e.m.f. is

A
Once per revolution

B
Twice per revolution

C
Four times per revolution

D
Six times per revolution

Solution

Consider the ring starting to rotate from its initial position where it is perpendicular to the magnetic field. Thus initially maximum flux is passing through the loop. When it rotates, the flux passing through the loop starts to decrease. When it becomes parallel to the magnetic field, the flux becomes zero and then starts to increase. Here the direction of the induced emf remains the same. Then as it rotates further, it keeps on increasing, reaches maximum. Here the direction of induced emf changes. According to lenz law, the induced emf tends to oppose the flux when it was increasing and now since it is decreasing, it tends to increase it.
Similarly when it reaches its initial position it changes the direction.