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Question: The magnetic field energy in an inductor changes from maximum value to minimum value in 5.0 ms when ...

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

Explanation

Solution

The magnetic energy stored in an inductor coil is given by the formula :
E=12LI2E = \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 ω\omega .

The voltage, V(t)=V0sin⁡ωtV\left( t \right) = {V_0}\sin \omega t
In a pure inductive circuit, the current delays by 90∘{90^ \circ } by the voltage.
The current, I(t)=I0sin⁡(ωt−π2)I\left( t \right) = {I_0}\sin \left( {\omega t - \dfrac{\pi }{2}} \right)
The energy stored in the inductor, E=12LI2E = \dfrac{1}{2}L{I^2}
Substituting, we get –
E=12LI02sin⁡2(ωt−π2)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 I0{I_0} and 0 in which cases the value ofsin⁡2(ωt−π2){\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⁡2ωt1−π2=1sin⁡2(ωt2−π2)=0{\sin^2}{\omega {t_1}} - \dfrac{\pi }{2} = 1 {\sin ^2}\left( {\omega {t_2} - \dfrac{\pi }{2}} \right) = 0
ωt1−π2=π2ωt2−π2=0\omega {t_1} - \dfrac{\pi }{2} = \dfrac{\pi }{2} \omega {t_2} - \dfrac{\pi }{2} = 0
ωt1=πωt2=π2\omega {t_1} = \pi \omega {t_2} = \dfrac{\pi }{2}
By subtracting the maximum and minimum values of phase, we get –
ωt1−ωt2=π−π2\omega {t_1} - \omega {t_2} = \pi - \dfrac{\pi }{2}
Taking ω\omega common and substituting from the given data that, (t1−t2)=5ms=5×10−3sec⁡\left( {{t_1} - {t_2}} \right) = 5ms = 5 \times {10^{ - 3}}\sec ,we get –
ω(t1−t2)=π2 →ω(5×10−3)=π2  \omega \left( {{t_1} - {t_2}} \right) = \dfrac{\pi }{2} \\\ \to \omega \left( {5 \times {{10}^{ - 3}}} \right) = \dfrac{\pi }{2} \\\
The angular frequency , ω=2πf\omega = 2\pi f where f is the frequency in hertz (Hz)
Thus,

2πf(5×10−3)=π2 →2πf(5×10−3)=π2 →f=12×2×5×10−3=10320 →f=100020=50Hz  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.