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 :
$$\frac{\pi }{4}\sqrt {LC} $$
Solution :
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 $$t = \frac{\pi }{4}\sqrt {LC} $$
Releted MCQ Question on Electrostatics and Magnetism >> Electromagnetic Induction
Releted Question 1
A thin circular ring of area $$A$$ is held perpendicular to a
uniform magnetic field of induction $$B.$$ $$A$$ small cut is made in the ring and a galvanometer is connected across the ends such that the total resistance of the circuit is $$R.$$ When the ring is suddenly squeezed to zero area, the charge flowing through the galvanometer is
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
Two identical circular loops of metal wire are lying on a table without touching each other. Loop-$$A$$ carries a current which increases with time. In response, the loop-$$B$$
A coil of inductance $$8.4 mH$$ and resistance $$6\,\Omega $$ is connected to a $$12 V$$ battery. The current in the coil is $$1.0 A$$ at approximately the time