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?
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
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:
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 -
$$L = \frac{{{\mu _0}{N^2}\pi {r^2}}}{\ell }$$
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
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