Question

A small circle loop of wire of radius a is located at the centre of a much larger circular wire loop of radius $b$. The two loops are in the same plane. The outer loop of radius $b$ carries an alternating current $I+I_{o}cos\left(\omega t\right)$ . The emf induced in the smaller inner loop is nearly :

• A. $\frac{\pi {\mu }_o{I}_o{b}^2}{a}\omega cos\left(\omega t\right)$
• B. $\pi {\mu }_o{I}_o\frac{a^2}{b}\omega sin\left(\omega t\right)$
• C. $\frac{\pi {\mu }_o{I}_o}{2}.\frac{a^2}{b}\omega sin\left(\omega t\right)$
• D. $\frac{\pi {\mu }_o{I}_o}{2}.\frac{a^2}{b}\omega cos\left(\omega t\right)$

Answer 1

Answer: (C)

$\frac{\pi {\mu }_o{I}_o}{2}.\frac{a^2}{b}\omega sin\left(\omega t\right)$

Magnetic field due to outer current loop

$B_o=\frac{{\mu }_{0}1}{2R}=\frac{{\mu }_0{1}_{0}cos\omega t}{2b}$

Induced emf in inner loop

$V=\frac{Nd\left(BA\right)}{dl}=1x\frac{d}{dl}\left(\frac{{\mu }_0{I}_{0}cos\omega t}{2b}x\pi a^2\right)$

$=\frac{\pi {\mu }_0{I}_0{a}^2}{2b}\frac{d}{dl}\left(cos\omega t\right)$

$=\frac{\pi {\mu }_0{I}_0{a}^2}{2b}\omega (-sin\omega t)$

$=\frac{\pi {\mu }_0{I}_0{a}^2}{2b}\omega sin\left(\omega t\right)$.