Question

a.Calculate the inductance of an air core solenoid containing $300$ turns if the length of the solenoid is $25.0cm$ and its cross- sectional area is $4.00c{m^2}$
b. Calculate the self -induced emf in the solenoid if the current through it is decreasing at the rate of $50A{s^{ - 1}}$

Answer 1

Hint : To calculate the inductance of the solenoid we use the formula $L = \dfrac{{{\mu _0}{N^2}S}}{l}$ .
Here everything is given, the turns that is $300$ , length of the solenoid, the cross sectional area and ${\mu _0}$ . We can calculate the inductance from these values.
Now to calculate the self-induced emf we put the obtained inductance value and the decreased rate of current.
$L = \dfrac{{{\mu _0}{N^2}S}}{l}$ for calculating inductance.
$e = - L\dfrac{{di}}{{dt}}$ For induced emf.

Complete Step By Step Answer:
In order to solve this question first we see what actually we have,
We have;
${\mu _0}$ Value as $4\pi \times {10^{ - 7}}$ , the number of turns of the coil which is three hundred, length of the solenoid which is $25.0cm$ and the cross sectional area of $4.00c{m^2}$ .
Now convert the length of the solenoid into meters, we get $25 \times {10^{ - 2}}$ meters.
Now according to the inductance of the solenoid,
L = \dfrac{{{\mu _0}{N^2}S}}{l} \\\ L = \dfrac{{(4\pi \times {{10}^{ - 7}}){{(300)}^2}\left( {4 \times {{10}^{ - 4}}} \right){N^2}S}}{{25 \times {{10}^{ - 2}}}} \\\ = 1.81 \times {10^{ - 4}}H \\\
Now we get the value of inductance of solenoid, to get the self- induced emf we write;
$e = - L\dfrac{{di}}{{dt}}$
$e$ Is the self-induced emf of the solenoid and
Here, $\dfrac{{di}}{{dt}}$ is the rate of change of current which is decreasing.
$\dfrac{{di}}{{dt}} = - 50A{s^{ - 1}}$
Now;
E = - (1.81 \times {10^{ - 4}})(50) \\\ = 9.05 \times {10^{ - 3}}V \\\ e = 9.05mV \\\
So induced emf is $9.05mV$ .

Note :
In this question first see the units and convert the centimeters into meter units. Use the proper formula with everything at the appropriate unit. To calculate the self- induced emf we put the obtained inductance value and the decreased rate of current.