Question

A plane loop is shaped in the form as shown in figure with radii $a=20\text{cm}$ and $b=10\text{cm}$ and is placed in a uniform time varying magnetic field $B=B_0\sin \omega t$ , where $B_0=10\text{mT}$ and $\omega =100\text{rad/s}$. Find the amplitude of the current induced(in $A$) in the loop, if its resistance per unit length is equal to $50\times {10}^{-3}\frac{\Omega }{\text{m}}$. The inductance of the loop is negligible.

Answer 1

If the magnetic field is increasing, flux in both loops are increasing, so that direction of current is anticlockwise but this wire is continuous so single current will flow.
Here $B=B_0\sin \omega t$ where $B_0=10\text{mT}$
$\omega =100\frac{rad}{s}\frac{R}{l}=50\times {10}^{-3}\frac{\Omega }{m}$
$a=20\text{cm},b=10\text{cm}$

Flux in bigger loop
${\phi }_1=B\pi a^2=B_0\pi a^2\sin \omega t$
$e_1=\frac{d{\varphi }_1}{dt}=B_0\pi a^2\omega \cos \omega t$
Flux in smaller loop ${\phi }_2=B.\pi b^2=B_0\pi b^2\sin \omega t$
$e_2=\frac{d{\varphi }_2}{dt}=B_0\pi b^2\omega \cos \omega t$
$E_{net}=e_1-e_2=\pi B_0\omega (a^2-b^2)cos\omega t$
Total resistance of wire $R=2\pi (a+b)\lambda$ induced current
$I=\frac{e}{R}=\frac{\pi B_0\omega (a^2-b^2)cos\omega t}{2\pi \lambda (a+b)}$
$I=\frac{B_0\omega (a-b)cos\omega t}{2\lambda }$
$I$ will be maximum when $\cos \omega t=1$
Current amplitude $I_0=\frac{B_0\omega (a-b)}{2\lambda }$
Put the values here
$I_0=\frac{10\times {10}^{-3}\times 100\times (20-10){10}^{-2}}{2\times 50\times {10}^{-3}}$
$I_0=1\text{A}$