Question

If the magnetic field energy stored in an inductor changes from maximum to minimum value in 5 ms, when connected to an a.c. source, the frequency of the a.c. source is:

• A. 200 Hz
• B. 500 Hz
• C. 50 Hz
• D. 20 Hz
• E. 100 Hz

Hint:

The energy stored in an inductor in an a.c. circuit changes with the frequency of the source. For energy to change from maximum to minimum, it takes half a cycle of the oscillation.

Answer 1

The Correct Option is C

Step 1: Understanding Magnetic Field Energy in an Inductor The magnetic energy stored in an inductor is given by: \[ U = \frac{1}{2} L I^2 \] where \( L \) is the inductance and \( I \) is the current. Since the circuit is connected to an AC source, the current varies sinusoidally as: \[ I = I_0 \sin(\omega t) \] where \( \omega = 2\pi f \) is the angular frequency. Thus, the energy stored in the inductor is: \[ U = \frac{1}{2} L I_0^2 \sin^2(\omega t) \] Step 2: Time for Energy to Change from Maximum to Minimum The energy \( U \) reaches its maximum when \( \sin^2(\omega t) = 1 \) and minimum when \( \sin^2(\omega t) = 0 \). This change occurs in a time interval equal to one-quarter of the time period \( T \) of the AC source: \[ \Delta t = \frac{T}{4} \] Given that this time is 5 ms: \[ \frac{T}{4} = 5 \times 10^{-3} { s} \] Step 3: Calculating the Frequency The total time period of the AC source is: \[ T = 4 \times (5 \times 10^{-3}) = 20 \times 10^{-3} { s} = 0.02 { s} \] The frequency is given by: \[ f = \frac{1}{T} = \frac{1}{0.02} = 50 { Hz} \] Thus, the correct answer is \( 50 \) Hz.