Question

The magnetic field energy in an inductor changes from maximum value to minimum value in $5.0ms$, 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

The correct option is B 50 Hz

Let inductor$=L,$ instantaneous current $=I\left(t\right)=I_{peak}\sin \left(\omega t\right)$

The magnetic field energy in an inductor is given by

$\frac{1}{2}L{I}^2\left(t\right)=\frac{1}{2}L{I}_{peak}^2{\sin }^2\left(\omega t\right)$

the energy goes from a maximum value $($ at time $t_1)$ to the next minimum value $($at time $t_2)$ when

$\omega t_1=\frac{\pi }{2}\&\omega t_2=\pi$

Or $\omega t_1=\frac{3\pi }{2}\&\omega t_2=2\pi$

In either case,

$\omega (t_2-t_1)=\frac{\pi }{2}$
$\omega \times 5\times {10}^{-3}=\frac{\pi }{2}$
$2f\times 5\times {10}^{-3}=\frac{1}{2}$
$f=\frac{{10}^3}{2}\times 2\times 5$
$f=50Hz$