A long cylindrical solenoid with 100 turns/cmhas a radius of 1.6 cm. Assume that the magnetic field it produces is parallel to its axis and is uniform in its interior. (a) What is its inductance per meter of length? (b) If the current changes at the rate of 13A/s, what emf is induced per meter?

$\mathrm{n}=100\mathrm{turns}/\mathrm{cm}=100\times {10}^2\mathrm{turns}/\mathrm{m}$

$r=1.6cm=1.6\times {10}^{-2}m$

The magnetic field it produces is parallel to its axis and is uniform in its interior.

Rate of current change, $\frac{di}{dt}=13A/s$

The inductance per unit length near the middle of a long solenoid of cross-sectional area and turns per unit length is given by equations 30-31. After that, we can use the relation between the emf and inductance of that coil to find the emf per unit length.

Formula:

$$\begin{gathered} \frac{L}{I}={\mu }_0{n}^{2}A \\ \epsilon =L\frac{di}{dt} \end{gathered}$$

As the magnetic field it produces is parallel to its axis and is uniform in its interior, the inductance per unit length can be written by using equation 30-31

As
$\frac{L}{I}={\mu }_0{n}^{2}A$

$\frac{L}{I}={\mu }_0{n}^2\pi r^2$

So, by putting the value,

$\frac{L}{I}=4\pi \times {10}^{-7}\times (100\times {10}^2{)}^2\times \pi \times (1.6\times {10}^{-2}{)}^2$

$\frac{L}{I}=4\pi \times {10}^{-7}\times {10}^4\times \pi \times 2.25$

$\frac{L}{I}=0.10H/m$

For inductor the emf is defined as

$\epsilon =L\frac{di}{dt}$

So that the $emf$per unit length is

$\frac{\epsilon }{I}=\frac{L}{I}\frac{di}{dt}$

By putting the value of the L/I and $\frac{di}{dt}=13A/s$, we get

$$\begin{gathered} \frac{\epsilon }{I}=0.10\times 13 \\ \frac{\epsilon }{I}=1.3V/m \end{gathered}$$