Question

Two concentric circular loops of radii $10cm$ and $100cm$ are placed on a horizontal plane as shown in the Figure. Outer loop is connected to a source of unknown emf through a key K. lf a current of $2A$ flows in the inner loop as shown aside, when the key k is closed at time $t=0s$, then the emf of the source sending a current $I$, should be

[Take resistance of both wire per length as ${10}^{-4}\Omega m^{-1}$)

• A. $4V$
• B. $4tV$
• C. $04t^2\vee$
• D. $(0.4t^2+4t+4)V$

Answer 1

Answer: (B)

$4tV$

Let V be the voltage supplied sending a current I in outer loop.

Resistance of outer loop $R_1=2\pi r_1\times {10}^{-4}=2\pi {10}^{-4}\Omega$

Current I in outer loop:$I=\frac{V}{2\pi \times {10}^{-4}}=\frac{V}{2\pi }\times {10}^{4}A$

Magnetic field B at the centre O: $B=\frac{{\mu }_{0}V\times {10}^4}{4\pi \times 1}$
$=\frac{4\pi \times {10}^{-7}V\times {10}^4}{4\pi }$
$=V\times {10}^{-3}T$

The flux associated with inner coil $\varphi =\pi r_2^2\times V\times {10}^{-3}=\pi V\times {10}^{-5}Wb$

$\therefore$ induced emf $|e|=\frac{d\varphi }{dt}=\pi \times {10}^{-5}\frac{dV}{dt}=R_2\times I_1$
Where the resistance of inner coil $R_2=2\pi r_2\times {10}^{-4}=2\pi {10}^{-5}\Omega$

$\therefore \pi \times {10}^{-5}\frac{dV}{dt}=4\pi \times {10}^{-5}$

$or,dV=4dt$
giving, $V=4t+C$
As $V=0$ at $t=0$
$V=4tV$
As $I_1$ is in clockwise direction, producing magnetic field into the plane, the current $I$ should be in anticlockwise direction, which is the cause for induction