Electromagnetic Induction Test

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Electromagnetic Induction Test - 53

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Question 1
1 / -0

A uniform rod of resistance $10\Omega$ is bent into a ring of radius $20\ cm$ . The ring is placed in a uniform magnetic field of induction $2.0T$ directed parallel to its plane. If ideal battery of emf $5.0\ V$ is connected between two points of the ring to obtain maximum force on the ring, then the value of this maximum force is

A
$0.4\ N$

B
$0.8\ N$

C
$1.2\ N$

D
$1.6\ N$

Solution
Question 2
1 / -0

An e.m.f. of 5 volt is produced by a self inductance, when the current changes at a steady rate from 3 A to 2 A in 1 millisecond. The value of self inductance is

A
zero

B
5 H

C
5000 H

D
5 mH

Solution

Hint:- Use formula of self emf induced in inductor

Step 1: Note the given values
$Emf induced, e=5v$
Change in current dI= (2-3) A=-1A
$Time,dt=1ms$
$Inductace,L=?$

Step 2: Calculate self inductance
we know that em.f develop by the inductor due to change in current
$E=-L\dfrac{\mathrm{dI} }{\mathrm{d} t}=-L\dfrac{(2-3)}{10^{-3}}=L\times10^3$
$\Rightarrow L\times10^3=5$
$\Rightarrow L=5mH$
Hence the correct option is D
Question 3
1 / -0

A metal disc rotates freely, between the poles of a magnet in the direction indicated. Brushes P and Q make contact with the edge of the disc and the metal axle. What current, if any, flows through R ?

A
a current from P to Q

B
a current from Q to P

C
no current, because the emf in the disc is opposed by the back emf

D
no current, because the emf induced in one side of the disc is opposed by the emf induced in the other side.

E
no current , because no radial emf is induced in the disc.

Solution

Given,

The disc is rotating in magnetic field from North to South.

If the direction of the magnetic field is perpendicular to the plane of the disc:
Viewing disk from the top.

Force on each proton is toward the inside, and force on each electron is outside.

So, the negative charge will develop, in the outskirt of the disk, and the positive charge will develop at the center.

Hence, Current will flow from P to Q.

$Alternatively:$
If the magnetic field is perpendicular to the normal vector of the plane of the disc:
The emf induced will not be radial, as governed by flaming's right hand rule, thus no current will flow through the resistance R.
Question 4
1 / -0

Two concentric co-planar circular loops of radii $r_1$ and $r_2$ carry currents of respectively $i_1$ and $i_2$ in opposite directions (one clockwise and the other anticlockwise.) The magnetic induction at the centre of loops is half that due to $i_1$ alone at the centre. If $r_2 = 2r_1$ . the value of $i_2 / i_1$ is

A
$2$

B
$\dfrac{1}{2}$

C
$\dfrac{1}{4}$

D
$1$

Solution

$\begin{array}{l} B=\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } -\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 2\left( { 2{ r_{ 1 } } } \right) } } \\ =\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } -\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 4{ r_{ 2 } } } } \\ { B^{ ' } }=\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } \\ \Rightarrow B=\frac { { { B^{ ' } } } }{ 2 } \\ \Rightarrow \frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 2{ r_{ 1 } } } } -\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 4{ r_{ 1 } } } } =\frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 4{ r_{ 1 } } } } \\ \Rightarrow \frac { { { \mu _{ o } }{ i_{ 1 } } } }{ { 4{ r_{ 1 } } } } =\frac { { { \mu _{ o } }{ i_{ 2 } } } }{ { 4{ r_{ 1 } } } } \\ \Rightarrow \frac { { { i_{ 1 } } } }{ { { i_{ 2 } } } } =1 \\ Hence, \\ option\, \, D\, \, is\, \, correct\, \, answer. \end{array}$
Question 5
1 / -0

A resistanceless conduct moves in uniform magnetic field on a uniform rectangular loop of resistance $R$ with constant velocity v moving from $CD$ to $EF$ . Current through conduct

A
Increases and then decreases

B
decreases and then increases

C
Remains constant

D
Decreases continuously

Solution
Question 6
1 / -0

Two inductance's connected in parallel are equivalent to a single inductance of $1.5 \ H$ , and when connected in series are equivalent to a single inductance of $8 \ H$ . The difference in their inductance is:-

A
$3 \ H$

B
$7.5 \ H$

C
$2 \ H$

D
$4 \ H$

Solution

D. $4 H$
Inductance in series combination,
$L= L_{1}+L_{2}$
$8= L_{1}+L_{2}$
Inductance in parallel combination,
$\dfrac{1}{L}$ $= \dfrac{1}{L_{1}}$ + $\dfrac{1}{L_{2}}$

$\dfrac{1}{1.5}$ $=\dfrac{L_{1}+L_{2}}{L_{1}.L_{2}}$

$\dfrac{1}{1.5}$ $=\dfrac{8}{L_{1}.L_{2}}$

$L_{1}.L_{2}= 12$

$L_{1}=$ $\dfrac{12}{L_{2}}$
Put value of $L_{1}$ in the series combination
$\dfrac{12}{L_{2}}+L_{2}= 8$
$L_{2}^2-8L_{2}+12=0$
$L_{2}= 2, 6$

$L_{1}=$ $\dfrac{12}{L_{2}}$
$=6,2$
So, the difference between the two inductance is 4.
Question 7
1 / -0

A long magnet moves with constant velocity along the axis of a fixed metal ring. It starts from a large distance from the ring, passes through the ring and then moves away far from the ring. The current in the ring is plotted against time. Which of the following best represents the resulting curve?

A

B

C

D

Solution

As the bar magnet passes through the coil, the flux linked with the ring first increases and then decreases. The current in the ring will therefore reverse in direction as the magnet passes through it. Also, the rate of change of flux is maximum when the magnet is close to the coil therefore the magnitude of current when the magnet is near coil is maximum.
Question 8
1 / -0

Flux linked through following coils changes with respect to time then for which coil an e.m.f. is not induced:

A
Copper coils

B
Wood coils

C
Iron coils

D
None

Solution
Question 9
1 / -0

A conducting disc of radius R is rotating with angular velocity $\omega$ .Mass of electron is and charged e. If electrons are the current carries in a conductor, the potential difference between the center and the edge of the disc is:

A
$\dfrac{{m{\omega ^2}{R^2}}}{e}$

B
$\dfrac{{m{\omega ^2}{R^2}}}{{4e}}$

C
$\dfrac{{m{\omega ^2}{R^2}}}{{3e}}$

D
$\dfrac{{m{\omega ^2}{R^2}}}{{2e}}$

Solution

Due to rotations, the electron experfence a centripctual force $F_{e}$ given by :-
$F_{e} = m\omega^{2}r$
Due to the fonce, a radical electric field is
generated which is :-
$E = \dfrac{F_{e}}{e} \,\,\,\, \left [ as F = qE \right ]$
$\Rightarrow E = \dfrac{m\omega^{2}r}{e}$
So, the potential difference generated between the centre
(r = 0) and edge of dise (r = R) is obtained as :-
V = $\displaystyle \int E. dr = \int_{o}^{R} \dfrac{m\omega^{2}r}{e}dr = \dfrac{m\omega^{2}}{e} \int_{o}^{R} r dr = \dfrac{m\omega^{2}}{e} \left [ \dfrac{r^{2}}{2} \right ]_{o}^{R}$
$\Rightarrow V = \dfrac{m\omega^{2}R^{2}}{2l}$
Question 10
1 / -0

According to the faraday's laws of electro magnetic induction-

A
The direction of the induced current is such that it opposes it self.

B
The induced emf in the coil is equal to the rate of change of magnetic flux associated with it.

C
The direction of induced emf is such that it opposes self.

D
None of the above

Solution