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.