Question

Figure shows two parallel and coaxial loops. The smaller loop (radius $r$) is above the larger loop (radius $R$), by distance $x$ ($x>>R$). The magnetic field due to current $i$ in the larger loop is nearly constant throughout the smaller loop. Suppose that $x$ is increasing at a constant rate of $\frac{dx}{dt}=v$. The induced emf in the smaller loop is

• A. $\frac{{\mu }_0\pi i{R}^2{r}^{2}v}{x^4}$
• B. $\frac{{\mu }_0\pi i{R}^2{r}^{2}v}{2x^4}$
• C. $\frac{3{\mu }_0\pi i{R}^2{r}^{2}v}{2x^4}$
• D. $\frac{{\mu }_0\pi i{R}^2{r}^{2}v}{3x^4}$

Answer 1

The correct option is C $\frac{3{\mu }_0\pi i{R}^2{r}^{2}v}{2x^4}$


Magnetic field due to larger loop
$B=\frac{{\mu }_{0}i{R}^2}{2(x^2+R^2{)}^{3/2}}$
Since $x>>R$, $B$ will be parallel to area vector of smaller loop. The flux through the smaller loop is
$\varphi =BA\cos \theta$
$\varphi =\frac{{\mu }_{0}i{R}^2\left(\pi r^2\right)cos0}{2(x^2+R^2{)}^{3/2}}$
$\varphi =\frac{{\mu }_{0}i{R}^2\left(\pi r^2\right)}{2(x^2+R^2{)}^{3/2}}$.........(1)
$R$ can be neglected in front of $x$
So (1) becomes
$\varphi =\frac{{\mu }_{0}i{R}^2\pi r^2}{2x^3}$

We know that,

$ϵ=-\frac{d\varphi }{dt}$
$\frac{d\varphi }{dt}=\frac{{\mu }_{0}i\pi R^2{r}^2}{2}\frac{d}{dt}\left(\frac{1}{x^3}\right)$
$\frac{d\varphi }{dt}=\frac{{\mu }_{0}i\pi R^2{r}^2}{2}\left(\frac{-3}{x^4}\right)\left(\frac{dx}{dt}\right)$
As $\frac{dx}{dt}=v$
$ϵ=-\frac{d\varphi }{dt}=\frac{3{\mu }_0\pi i{R}^2{r}^{2}v}{2x^4}$