Electromagnetic Induction Test

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Electromagnetic Induction Test - 55

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Question 1
1 / -0

The time constant of a circuit is 10 sec, When a resistance of $$ 100 \Omega $$ is connected in series in a previous circuit then time constant becomes 2 second,then the self inductance of the circuit is;-

A
$$250 H$$

B
$$50H$$

C
$$150 H$$

D
$$25 H$$

Solution

In LR circuit,
The time constant $$\tau=\dfrac{L}{R}$$
$$10=\dfrac{L}{R}$$
$$L=10R$$. . . . . . .(1)
When Resistance $$100\Omega $$ is connect in series, than the time constant is
$$\tau'=\dfrac{L}{R+100}=2s$$
$$L=2R+200$$. . . . . . .(2)
Equating equation (1 ) and (2), we get
$$2R+200=10R$$
$$8R=200$$
$$R=25\Omega$$
From equation (1),
$$L=10R=10\times 25$$
$$L=250H$$
Question 2
1 / -0

A rectangular loop of sides 'a' and 'b' is placed in the XY plane. A very long wire is also placed in xy plane such that side of length 'a' of the loop is parallel to the wire. The distance between the wire and the nearest edge of the loop is 'd'. The mutual inductance of this system is proportional to?

A
a

B
b

C
$$1/d$$

D
Current in wire
Question 3
1 / -0

A conducting rod AB moves parallel to x-axis in a uniform magnetic field, pointing in the positive z-direction. The end A of the rod gets positively charged. Is this statement true?

A
Yes

B
No

C
Not defined

D
Any answer is right

Solution
Question 4
1 / -0

There is a uniform (in spatial districution) magnetic filed B in a circular region of radius R as shown in the figure whose magnitude varies uniformly a the rate $$\beta$$ w.r.t. time. The emf induced across the ends of a circular concentric conducting arc of radius $$R_1$$ having an agle $$\theta$$ as shown $$( < OAO'=\theta)$$ is

A
$$\dfrac{\theta}{2 \pi}R^2_1. \beta$$

B
$$\dfrac{\theta}{2 }R^2. \beta$$

C
$$\dfrac{\theta}{2 \pi}R^2. \beta$$

D
Zero

Solution

$$Emf = \dfrac{d \phi}{dt}$$
where
$$\dfrac{dB}{dt} = \beta$$
$$\phi = \displaystyle \int B. ds$$
$$= B \dfrac{1}{2} r^2 . \dfrac{\theta}{\pi}$$
$$= \dfrac{Br^2 \theta}{2 \pi}$$
$$Emf = \dfrac{dB}{dt} \dfrac{r^2 \theta}{2 \pi}$$
$$= \dfrac{\beta r^2 \theta}{2 \pi} $$
$$Emf = \dfrac{\theta}{2 \pi} r^2 . \beta$$
Question 5
1 / -0

An electron beam is moving near to a conducting loop then the induced current in the loop:-

A
clockwise

B
anticlockwise

C
both

D
no current

Solution

The induced current in a loop moves from right to left. Since the electrons flow in direction opposite to that of the current thus the electron beam will move from left to right (anticlockwise).
Question 6
1 / -0

A square conducting loop, whose plane is perpendicular to a uniform magnetic field, moves with velocity $$v$$ normally to the magnetic field. If opposite sides of the loop, perpendicular to its velocity, remain in two mutually opposite uniform magnetic fields of strength $$B$$, the induced electromotive force in this coil will be ______ . Length of each side is $$l$$.

A
$$2Bvl$$

B
$$Bvl$$

C
$$0$$

D
$$\dfrac{Bvl}{2}$$

Solution
Question 7
1 / -0

A point charge of $$0.1\ C$$ is placed on the circumference of a non-conducting ring of radius $$1m$$ which is rotating with a constant angular acceleration of $$1\ rad/sec^{2}$$. If ring starts it's motion from rest at $$t = 0$$, the magnetic field at the centre of the ring at $$t =10\ sec$$, is

A
$$10^{-6}T$$

B
$$10^{-7}T$$

C
$$10^{-8}T$$

D
none

Solution
Question 8
1 / -0

An atom is placed in a uniform magnetic induction $$B$$ such that plane normal of the electron orbit makes an angle of $$30^\circ$$ with the magnetic induction. The torque acting on the orbiting electron is:-

A
$$\dfrac{ehB}{8 \pi m}$$

B
$$\dfrac{ehB}{4 \pi m}$$

C
$$\dfrac{2ehB}{3 \pi m}$$

D
zero

Solution
Question 9
1 / -0

In a long hollow vertical metal pipe a magnet is dropped. During its fall, the acceleration of magnet

A
will decrease linearly

B
will decrease upto a value which is less than $$g$$.

C
will decrease to zero and will attain a terminal speed

D
may increase or decrease

Solution

The falling magnet will experience an upward force and
so the acceleration of magnet will decreases and then the magnet will a quire a terminal speed.
Hence $$C$$ is the correct answer.
Question 10
1 / -0

A circular coil has $$500$$ turns of wire and its radius is $$5cm$$. The self inductance of the coil is:-

A
$$25 \times 10^{-3} mH$$

B
$$25 mH$$

C
$$50 \times 10^{-3} H$$

D
$$50 \times 10^{-3} mH$$

Solution

$$\varphi = Li $$
$$\Rightarrow NBA=Li$$ Since magnetic field at the center of circular coil
carring current is given by $$B = \dfrac{{{\mu _0}}}{{4\pi }}.\dfrac{{2\pi Ni}}{r}$$
$$\therefore$$ $$N.\dfrac{{{\mu _0}}}{{4\pi }}.\dfrac{{2\pi Ni}}{r} = Li$$
$$ \Rightarrow L = \dfrac{{{\mu _0}{N^2}\pi r}}{2}$$
Hence,
self conductance of coil is
$$ = \dfrac{{4\pi \times 10 \times 500 \times 500 \times \pi \times 0.05}}{2}$$
$$=25mH$$
Hence,
self conductance of coil is $$25mH$$
So option $$B$$ is correct answer.