91影视

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

$r=1.6cm=1.6脳{10}^{-2}m$

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:

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

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
$\frac{L}{I}={渭}_0{n}^{2}A$

$\frac{L}{I}={渭}_0{n}^2蟺r^2$

So, by putting the value,

$\frac{L}{I}=4蟺脳{10}^{-7}脳(100脳{10}^2{)}^2脳蟺脳(1.6脳{10}^{-2}{)}^2$

$\frac{L}{I}=4蟺脳{10}^{-7}脳{10}^4脳蟺脳2.25$

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

For inductor the emf is defined as

$蔚=L\frac{di}{dt}$

So that the $emf$per unit length is

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

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

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