Electromagnetic Induction MCQ Questions & Answers in Electrostatics and Magnetism | Physics
Learn Electromagnetic Induction MCQ questions & answers in Electrostatics and Magnetism are available for students perparing for IIT-JEE, NEET, Engineering and Medical Enternace exam.
51.
A $$100\,mH$$ coil carries a current of $$1\,A.$$ Energy stored in its magnetic field is
A
$$0.5\,J$$
B
$$1\,A$$
C
$$0.05\,J$$
D
$$0.1\,J$$
Answer :
$$0.05\,J$$
Energy stored in coil is $$E = \frac{1}{2}L{i^2}$$
where, $$L$$ is self-inductance of coil and $$i$$ is current induced. Here, $$L = 100\,mH = 100 \times {10^{ - 3}}H$$ and $$i = 1\,A$$
$$\therefore E = \frac{1}{2} \times \left( {100 \times {{10}^{ - 3}}} \right) \times {\left( 1 \right)^2} = 0.05\,J$$
52.
In an inductor of self-inductance $$L = 2\,mH,$$ current changes with time according to relation $$i = {t^2}{e^{ - t}}.$$ At what time emf is zero?
53.
In a uniform and constant magnetic field of induction $$B,$$ two long conducting wires $$ab$$ and $$cd$$ are kept parallel to each other at distance $$\ell $$ with their plane perpendicular to $$B.$$ The ends $$a$$ and $$c$$ are connected together by an ideal inductor of inductance $$L.$$ A conducting slider wire $$PQ$$ is imparted a speed $${v_0}$$ at time $$t = 0.$$ The situation is shown in the figure.
At time $$t = \frac{{\pi \sqrt {mL} }}{{4B\ell }},$$ the value of current $$I$$ through the wire $$PQ$$ is (ignore any resistance, electrical as well as mechanical)
54.
A thin semi-circular conducting ring of radius $$R$$ is falling with its plane vertical in horizontal magnetic induction $$\overrightarrow B .$$ At the position $$MNQ$$ the speed of the ring is $$v,$$ and the potential difference developed across the ring is
A
zero
B
$$\frac{{Bv\pi {R^2}}}{2}$$ and $$M$$ is at higher potential
C
$$\pi RBv$$ and $$Q$$ is at higher potential
D
$$2RBv$$ and $$Q$$ is at higher potential
Answer :
$$2RBv$$ and $$Q$$ is at higher potential
Induced emf produced across $$MNQ$$ will be same as the induced emf produced in straight wire $$MQ.$$
$$\therefore e = Bv\ell = Bv \times 2R\,{\text{with }}Q{\text{ at higher potential}}{\text{.}}$$
55.
The current $$\left( I \right)$$ in the inductance is varying with time according to the plot shown in figure.
Which one of the following is the correct variation of voltage with time in the coil?
A
B
C
D
Answer :
For inductor, as we know induced voltage for
$$t = 0\,{\text{to}}\,t = \frac{T}{2},$$
$$V = L\frac{{dI}}{{dt}} = L\frac{d}{{dt}}\left( {\frac{{2{I_0}t}}{T}} \right) = {\text{constant}}$$
For $$t = \frac{T}{2}\,{\text{to}}\,t = T,$$
$$V = L\frac{{dI}}{{dt}} = \left( {\frac{{ - 2{I_0}t}}{T}} \right) = - {\text{constant}}$$
So, answer can be represented with graph (D).
56.
A wire loop is rotated in a magnetic field. The frequency of change of direction of the induced emf is
A
once per revolution
B
twice per revolution
C
four times per revolution
D
six times per revolution
Answer :
twice per revolution
If a wire loop is rotated in a magnetic field, the frequency of change in the direction of the induced emf is twice per revolution.
57.
A fully charged capacitor $$C$$ with initial charge $${q_0}$$ is connected to a coil of self inductance $$L$$ at $$t = 0.$$ The time at which the energy is stored equally between the electric and the magnetic fields is:
A
$$\frac{\pi }{4}\sqrt {LC} $$
B
$$2\pi \sqrt {LC} $$
C
$$\sqrt {LC} $$
D
$$\pi \sqrt {LC} $$
Answer :
$$\frac{\pi }{4}\sqrt {LC} $$
Energy stored in magnetic field = $$\frac{1}{2}L{i^2}$$
Energy stored in electric field = $$\frac{1}{2}\frac{{{q^2}}}{C}$$
$$\eqalign{
& \therefore \frac{1}{2}L{i^2} = \frac{1}{2}\frac{{{q^2}}}{C} \cr
& {\text{Also }}q = {q_0}\cos \omega t{\text{ and }}\omega = \frac{1}{{\sqrt {LC} }} \cr} $$
On solving $$t = \frac{\pi }{4}\sqrt {LC} $$
58.
A wire of fixed lengths is wound on a solenoid of length $$\ell $$ and radius $$r.$$ Its self inductance is found to be $$L.$$ Now if same wire is wound on a solenoid of length $$\frac{\ell }{2}$$ and radius $$\frac{r}{2},$$ then the self inductance will be -
A
$$2L$$
B
$$L$$
C
$$4L$$
D
$$8L$$
Answer :
$$2L$$
$$L = \frac{{{\mu _0}{N^2}\pi {r^2}}}{\ell }$$
Length of wire $$ = N\,2\pi r = {\text{constant}}\left( { = C,{\text{suppose}}} \right)$$
$$\eqalign{
& \therefore L = {\mu _0}{\left( {\frac{C}{{2\pi r}}} \right)^2}\frac{{\pi {r^2}}}{\ell } \cr
& \therefore L \propto \frac{1}{\ell } \cr} $$
$$\therefore $$ Self inductance will become $$2L.$$
59.
A long solenoid has $$500$$ turns. When a current of $$2$$ ampere is passed through it, the resulting magnetic flux linked with each turn of the solenoid is $$4 \times {10^{ - 3}}Wb.$$ The self- inductance of the solenoid is
A
2.5 henry
B
2.0 henry
C
1.0 henry
D
40 henry
Answer :
1.0 henry
Total number of turns in the solenoid, $$N = 500$$
Current, $$I = 2\,A.$$
Magnetic flux linked with each turn $$ = 4 \times {10^{ - 3}}Wb$$
As, $$\phi = LI\,{\text{or}}\,N\phi = LI$$
$$ \Rightarrow L = \frac{{N\phi }}{1} = \frac{{500 \times 4 \times {{10}^{ - 3}}}}{2}{\text{henry}} = 1\,H.$$
60.
Two coaxial solenoids of different radius carry current $$I$$ in the same direction. $$\overrightarrow {{F_1}} $$ be the magnetic force on the inner solenoid due to the outer one and $$\overrightarrow {{F_2}} $$ be the magnetic force on the outer solenoid due to the inner one. Then :
A
$$\overrightarrow {{F_1}} $$ is radially inwards and $$\overrightarrow {{F_2}} = 0$$
B
$$\overrightarrow {{F_1}} $$ is radially outwards and $$\overrightarrow {{F_2}} = 0$$
C
$$\overrightarrow {{F_1}} = \overrightarrow {{F_2}} = 0$$
D
$$\overrightarrow {{F_1}} $$ is radially inwards and $$\overrightarrow {{F_2}} $$ is radially outwards