Electromagnetic Induction Test

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Electromagnetic Induction Test - 29

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Weekly Quiz Competition

Question 1
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The electric flux through a certain area of dielectric is $$8.76 \times 10^3 t^4$$. The displacement current through the area is $$12.9 pA$$ at $$t = 26.1 ms$$ . Find the dielectric constant of the material.

A
$$2 \times 10^{-8}$$

B
$$4 \times 10^{-8}$$

C
$$8 \times 10^{-8}$$

D
$$2 \times 10^{-7}$$

Solution

Displacement current is $$\displaystyle i_D = \varepsilon \dfrac{d\phi_E}{dt}$$ or
$$\displaystyle \varepsilon = \dfrac{i_D}{\dfrac{(d\phi_E}{dt})}$$
$$\displaystyle \varepsilon = \dfrac{12.9 \times 10^{-9}}{4 (8.76) \times 10^3 \times (26.1 \times 10^{-3})^3}$$
$$\simeq 2 \times 10^{-8}$$
Question 2
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The armature of a demonstrator generator consists of a flat square coil of side $$4 cm$$ and $$200$$ turns. The coil rotates in a magnetic field of $$0.75 T$$. The angular speed so that a maximum emf of $$1.6 V$$ is generated is:

A
$$\displaystyle \dfrac{20}{3}$$ rad/s

B
$$\displaystyle \dfrac{10}{3}$$ rotations/s

C
$$\displaystyle \dfrac{40}{3}$$ rpm

D
None of these

Solution

Maximum emf of a rotating coil $$\varepsilon_{max} = NA_0 B \omega $$ or
$$\displaystyle \omega = \dfrac{\varepsilon_{max}}{NA_0 B}$$
$$=\displaystyle \dfrac{1.6}{200 \times 16 \times 10^{-4} (.75)}$$
$$= \displaystyle \dfrac{20}{3} rad/s$$
Question 3
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A wire bent as a parabola $$y = kx^2$$ is located in a uniform magnetic field of induction $$B$$, the vector $$B$$ being perpendicular to the plane xy. At $$t = 0$$, sliding wire starts sliding from the vertex O with a constant acceleration a linearly as shown in figure. Find the emf induced in the loop.

A
$$By\displaystyle \sqrt{\frac{2a}{k}}$$

B
$$By\displaystyle \sqrt{\frac{4a}{k}}$$

C
$$By\displaystyle \sqrt{\frac{8a}{k}}$$

D
$$By\displaystyle \sqrt{\frac{a}{k}}$$

Solution

$$d \phi = B.dA = 2B x . dy$$ and $$y = kx^2$$
$$\therefore \displaystyle x = \sqrt{\dfrac{y}{k}}$$
$$\therefore \displaystyle \epsilon = \dfrac{d\phi}{dt} = 2 B \sqrt{\dfrac{y}{k}} \dfrac{dy}{dt}$$
using $$v^2 = 2as$$
$$\displaystyle \dfrac{dy}{dt} = v = \sqrt{2ay}$$
In y direction
or $$ |\varepsilon| = \displaystyle \dfrac{d\phi}{dt} = 2 B \sqrt{\dfrac{y}{k}} \sqrt{2ay}$$
or $$\displaystyle |\varepsilon| = By \sqrt{\dfrac{8a}{k}}$$
Question 4
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Directions For Questions

Consider a uniform metallic disc rotating through a perpendicular magnetic field $$\vec B$$ as shown in figure. Mass of the disk is $$M$$. Its radius is $$R$$ and thickness $$t$$. It is made of a material of resistivity $$\rho$$, and is rotating clockwise as shown in figure with angular speed $$\omega$$.Magnetic field is directed into the plane of the disc. Assume the region to which magnetic field is confined is a square of side $$L (L << R) $$ centered a distance $$d$$ from point O, the center of the disk. The sides of this square are horizontal and vertical.

...view full instructions

Find the approximate value of induced current assuming the resistance to the current is confined to the square.

A
$$\displaystyle \dfrac{BL \omega dt}{\rho}$$

B
$$\displaystyle \dfrac{BL^2 \omega dt}{\rho}$$

C
$$\displaystyle \dfrac{BL^2 \omega d}{\rho}$$

D
$$\displaystyle \dfrac{BL \omega d^2}{\rho}$$

Solution

$$\displaystyle emf,\ \varepsilon =\dfrac{d \phi}{dt} = BL^2 \omega$$

$$\displaystyle I = \dfrac{\varepsilon}{R}= \frac{BL^2 \omega}{\rho \dfrac{L}{dt}}= \dfrac{BL \omega dt}{\rho}$$
Question 5
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A long wire carrying current $$i$$ is placed close to a U-shaped conductor (of negligible resistance). A wire of length $$l$$ as shown in figure slides with a velocity $$v$$. Find the current induced in the loop as a function of distance $$x$$ from the current carrying wire to slider.

A
$$\displaystyle \dfrac{\mu_0 ilu}{Rx}$$

B
$$\displaystyle \dfrac{\mu_0 ilu}{2 \pi Rx}$$

C
$$\displaystyle \dfrac{\mu_0 ilu}{2Rx}$$

D
$$\displaystyle \dfrac{\mu_0 ilu}{4 \pi Rx}$$

Solution

$$B = \displaystyle \dfrac{\mu_0 i}{2 \pi x}$$

$$\varepsilon = Blv = \displaystyle \dfrac{\mu_0 ilv}{2 \pi x}$$

$$ \displaystyle I = \dfrac{\varepsilon}{R} = \dfrac{\mu_0 ilu}{2 \pi Rx}$$
Question 6
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The self inductance of a motor of an electric fan is $$10 H$$. In order to impart maximum power at $$50 Hz$$, it should be connected to a capacitance of :

A
$$\displaystyle 4\mu F$$

B
$$\displaystyle 8\mu F$$

C
$$\displaystyle 1\mu F$$

D
$$\displaystyle 2\mu F$$

Solution

Maximum power is transferred at resonance.
$$\displaystyle \therefore f_0 = \dfrac{1}{2\pi \sqrt{LC}}$$
or $$\displaystyle C = \dfrac{1}{4\pi^2 f^2_0 L} = \dfrac{1}{4\times 10\times(50)^2\times 10}$$
$$\displaystyle = 10^{-6}F = 1\mu F$$
Question 7
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How many meters of a thin wire are required to design a solenoid of length $$1 m $$ and $$L = 1mH$$? Assume cross-sectional diameter is very small.

A
$$10 m$$

B
$$40 m$$

C
$$70 m$$

D
$$100 m$$

E
$$140 m$$

Solution

Length of the wire $$l = nl_0 2 \pi r$$ and $$L = \mu_0 n^2 l_0 \pi r^2$$
or $$\displaystyle n = \sqrt{\dfrac{L}{\mu_0 l_0 \pi r^2}}$$
Thus, $$\displaystyle l = \sqrt{\dfrac{L}{\mu_0 l_0 \pi r^2}} l_0 2 \pi r$$
$$ \displaystyle 2 \pi r = \sqrt{\dfrac{Ll_0 4 \pi}{\mu_0}}$$
$$ = \displaystyle \sqrt{\dfrac{10^{-3} \times 1 \times 4 \pi}{4 \pi \times 10^{-7}}} = 100 m$$
Question 8
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A copper wire of length $$l$$ is bent into a semicircle. It is moved with a velocity $$v$$ in a region where magnetic field is uniform and perpendicular to the plane of the wire. If the strength of the field is $$B$$ then emf induced is

A
$$Blv$$

B
$$B\displaystyle \dfrac{l}{\pi}v$$

C
$$B\displaystyle \dfrac{2l}{\pi}v$$

D
none of these

Solution

$$ \pi r=l\\
\varepsilon = B (2r) v = B \left ( \dfrac{2l}{\pi} \right )v $$
Question 9
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A rod AB moves with a uniform velocity $$v$$ in a uniform magnetic field as shown in figure :

A
The rod becomes electrically charged

B
The end A becomes positively charged

C
The end B becomes positively charged

D
The rod becomes hot because of Joule heating

Solution

Force on positive charges in the rod will be in direction given by $$ q \vec v \times \vec B$$, i.e., towards A.
Hence force on electron will be in opposite direction i.e., towards B. So negative charge will move towards B.
Therefore B will be negatively charged and A will be positively charged.
Question 10
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A conducting rod is moved with a constant velocity $$v$$ in a magnetic field. A potential difference appears across the two ends

A
if $$\vec v || \vec l$$

B
if $$\vec v || \vec B$$

C
if $$\vec l || \vec B$$

D
none of these

Solution

A potential difference appears across the two ends if a component of length of rod and the a component of velocity of the rod and a component of magnetic field all mutually perpendicular.
that is $$ (\vec l \times \vec v ) . \vec B \neq 0 $$