(a) (i) Define self-inductance of a coil. Derive the expression for the energy required to build up a current I in a coil of self-inductance L. (ii)…

$\displaystyle E=-L \frac{d I}{d t}$

$\displaystyle U=\frac{1}{2} L I^{2}$ — parabolic graph obtained.

($\displaystyle 1$ indicates $\displaystyle 10$ mH) & ($\displaystyle 2$ indicates $\displaystyle 20$ mH)

Mutual inductance is defined as the emf induced in the primary coil when the current in the secondary coil changes at unit rate.

Mutual inductance is defined as the magnetic flux linked with the primary coil when the current in the secondary coil is unity.

Consider two long co-axial solenoids each of length l. Radius of inner solenoid $\displaystyle S_{1}$ is $\displaystyle r_{1}$ and number of turns per unit length is $\displaystyle n_{1}$. The corresponding quantities for outer solenoid $\displaystyle S_{2}$ are $\displaystyle r_{2}$ and $\displaystyle n_{2}$ respectively. Let $\displaystyle N_{1}$ and $\displaystyle N_{2}$ be the total number of turns of coils $\displaystyle S_{1}$ and $\displaystyle S_{2}$ respectively.

When a current $\displaystyle I_{2}$ is set up through $\displaystyle S_{2}$, it sets up magnetic flux through $\displaystyle S_{1}$:

$\displaystyle N_{1} \phi_{1}=M_{12} I_{2}$

$\displaystyle =\left(n_{1} l\right) \times\left(\pi r_{1}^{2}\right) \times\left(\mu_{0} n_{2} I_{2}\right)$

$\displaystyle =\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l I_{2}$

$\displaystyle M_{12}=\mu_{0} n_{1} n_{2} \pi r_{1}^{2} l=M_{21}$

$\displaystyle e=L \frac{d I}{d t}$

$\displaystyle L=\frac{e}{\dfrac{d I}{d t}}$

$\displaystyle =\frac{5 \times 10^{-3}}{\dfrac{2}{40}}$

$\displaystyle =0.1 \mathrm{H}$

$\displaystyle \phi=L I$

$\displaystyle =0.1 \times \frac{2}{40} \times 10$

$\displaystyle =0.05 \mathrm{~Wb}$