Question

The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0 ms when connected to an AC source. The frequency of the source is:
A. 20 Hz
B. 50 Hz
C. 200 Hz
D. 500 Hz

Answer 1

Hint:The magnetic energy stored in an inductor coil is given by the formula :
$E = \dfrac{1}{2}L{I^2}$
where I = current flowing through the coil in ampere (A) and L = self-inductance of the coil in henry(H).

Complete step-by-step answer:
Consider an inductor of inductance L connected to a voltage source of instantaneous voltage V of angular frequency $\omega $.

The voltage, $V\left( t \right) = {V_0}\sin \omega t$
In a pure inductive circuit, the current delays by ${90^ \circ }$ by the voltage.
The current, $I\left( t \right) = {I_0}\sin \left( {\omega t - \dfrac{\pi }{2}} \right)$
The energy stored in the inductor, $E = \dfrac{1}{2}L{I^2}$
Substituting, we get –
$E = \dfrac{1}{2}L{I_0}^2{\sin ^2}\left( {\omega t - \dfrac{\pi }{2}} \right)$
The maximum and minimum values of the magnetic energy in this case are when values of current are ${I_0}$ and 0 in which cases the value of${\sin ^2}\left( {\omega t - \dfrac{\pi }{2}} \right)$ are 1 and 0 respectively.
The angles or phases at which they become 1 and 0 respectively, are given by :
${\sin^2}{\omega {t_1}} - \dfrac{\pi }{2} = 1 {\sin ^2}\left( {\omega {t_2} - \dfrac{\pi }{2}} \right) = 0 $
$ \omega {t_1} - \dfrac{\pi }{2} = \dfrac{\pi }{2} \omega {t_2} - \dfrac{\pi }{2} = 0 $
$ \omega {t_1} = \pi \omega {t_2} = \dfrac{\pi }{2} $
By subtracting the maximum and minimum values of phase, we get –
$\omega {t_1} - \omega {t_2} = \pi - \dfrac{\pi }{2}$
Taking $\omega $ common and substituting from the given data that, $\left( {{t_1} - {t_2}} \right) = 5ms = 5 \times {10^{ - 3}}\sec $,we get –
$
  \omega \left( {{t_1} - {t_2}} \right) = \dfrac{\pi }{2} \\
   \to \omega \left( {5 \times {{10}^{ - 3}}} \right) = \dfrac{\pi }{2} \\
 $
The angular frequency , $\omega = 2\pi f$ where f is the frequency in hertz (Hz)
Thus,
\[
  2\pi f\left( {5 \times {{10}^{ - 3}}} \right) = \dfrac{\pi }{2} \\
   \to 2\pi f\left( {5 \times {{10}^{ - 3}}} \right) = \dfrac{\pi }{2} \\
   \to f = \dfrac{1}{{2 \times 2 \times 5 \times {{10}^{ - 3}}}} = \dfrac{{{{10}^3}}}{{20}} \\
   \to f = \dfrac{{1000}}{{20}} = 50Hz \\
 \]
Therefore, the frequency, f = 50 Hz.

Hence, the correct option is Option B.

Note:The inductor works on the principle of electromagnetic induction. This is the principle on which the inductor is able to store the magnetic energy inside.
When there is changing current in the coil, the magnetic flux arising from that current changes. Whenever the flux changes, it sets up an emf called back emf in the coil, which opposes the change in the magnetic field inside the coil. This phenomenon is called self-inductance. Through this self-inductance, by passing current over time, the coil is able to store the magnetic energy.