Question

Two coils have a mutual inductance 0.005 H. The current changes in the first coil according to equation \( I = I_0 \sin \omega t \), where \( I_0 = 10A \) and \( \omega = 100\pi \, \text{radian/sec} \). The maximum value of e.m.f. in the second coil is:

• A. \( 2\pi \)
• B. \( 5\pi \)
• C. \( \pi \)
• D. \( 4\pi \)

Hint:

The maximum induced e.m.f. is proportional to the mutual inductance, the current amplitude, and the angular frequency.

Answer 1

The Correct Option is A

Step 1: The maximum e.m.f. induced in the second coil is given by the formula \( \mathcal{E} = M \dfrac{dI}{dt} \), where \( M \) is the mutual inductance and \( \dfrac{dI}{dt} \) is the rate of change of current.
Step 2: The current is given by \( I = I_0 \sin(\omega t) \), so \( \dfrac{dI}{dt} = I_0 \omega \cos(\omega t) \). The maximum value occurs when \( \cos(\omega t) = 1 \).
Step 3: The maximum e.m.f. is: \[ \mathcal{E_{\text{max}}} = M I_0 \omega = 0.005 \times 10 \times 100\pi = 2\pi \, \text{V}. \]
Final Answer: \[ \boxed{2\pi \, \text{V}} \]