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 $L={\displaystyle \frac{{\mu }_0{N}^{2}S}{l}}$ .
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.
$L={\displaystyle \frac{{\mu }_0{N}^{2}S}{l}}$ for calculating inductance.
 $e=-L{\displaystyle \frac{di}{dt}}$ 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,
$$\begin{gathered} L={\displaystyle \frac{{\mu }_0{N}^{2}S}{l}} \\ L={\displaystyle \frac{(4\pi \times {10}^{-7}){(300)}^2\left(4\times {10}^{-4}\right)N^{2}S}{25\times {10}^{-2}}} \\ =1.81\times {10}^{-4}H \end{gathered}$$
Now we get the value of inductance of solenoid, to get the self- induced emf we write;
  $e=-L{\displaystyle \frac{di}{dt}}$
 $e$ Is the self-induced emf of the solenoid and
Here, ${\displaystyle \frac{di}{dt}}$ is the rate of change of current which is decreasing.
 ${\displaystyle \frac{di}{dt}}=-50A{s}^{-1}$
Now;
$$\begin{gathered} E=-(1.81\times {10}^{-4})(50) \\ =9.05\times {10}^{-3}V \\ e=9.05mV \end{gathered}$$
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.