Question

Find the mutual inductance between the circular coil and the loop shown in figure.

Answer 1

The magnetic field due to coil 1 at the centre of coil 2 is given by
$B=\frac{{\mathrm{\mu }}_{0}Ni{a}^2}{2(a^2+x^2{)}^{3/2}}$

The flux linked with coil 2 is given by
$\varphi =B.A\text{'}=\frac{{\mathrm{\mu }}_{0}Ni{a}^2}{2(a^2+x^2{)}^{3/2}}\mathrm{\pi }a{\text{'}}^2$

Now, let y be the distance of the sliding contact from its left end.
Given:
$v=\frac{dy}{dt}$
Total resistance of the rheostat = R
When the distance of the sliding contact from the left end is y, the resistance of the rheostat is given by
$r\text{'}=\frac{R}{L}y$

The current in the coil is the function of distance y travelled by the sliding contact of the rheostat. It is given by
$i=\frac{E}{\left({\displaystyle \frac{R}{L}}y+r\right)}$

The magnitude of the emf induced can be calculated as:
$e=\frac{d\mathrm{\varphi }}{dt}=\frac{{\mu }_{0}N{a}^{2}a{\text{'}}^2\mathrm{\pi }}{2(a^2+x^2{)}^{3/2}}\frac{di}{dt}$

$$\begin{gathered} e=\frac{{\mathrm{\mu }}_{0}N\mathrm{\pi }{a}^{2}a{\text{'}}^2}{2(a^2+x^2{)}^{3/2}}\frac{d}{dt}\frac{E}{\left({\displaystyle \frac{R}{L}}y+r\right)} \\ e=\frac{{\mathrm{\mu }}_{0}N\mathrm{\pi }{a}^{2}a{\text{'}}^2}{2(a^2+x^2{)}^{3/2}}\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^2}\right] \end{gathered}$$

emf induced, $e=\frac{{\mathrm{\mu }}_{0}N\mathrm{\pi }{a}^{2}a{\text{'}}^2}{2(a^2+x^2{)}^{3/2}}\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^2}\right]$
The emf induced in the coil can also be given as:
$\frac{\mathrm{d}i}{\mathrm{d}t}=\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^2}\right]$
$$\begin{gathered} e=M\frac{\mathrm{d}i}{\mathrm{d}t},\frac{\mathrm{d}i}{\mathrm{d}t}=\left[E\frac{\left(-{\displaystyle \frac{R}{L}}v\right)}{{\left({\displaystyle \frac{R}{L}}y+r\right)}^2}\right] \\ M=\frac{e}{{\displaystyle \frac{\mathrm{d}i}{\mathrm{d}t}}}=\frac{N{\mu }_0\mathrm{\pi }{a}^{2}a{\text{'}}^2}{2(a^2+x^2{)}^{3/2}} \end{gathered}$$