Question

A coil of conducting wire carries a current of i(t) = 14.0 sin(1.15 ✕ 103t), where i is in amperes and t is in seconds. A second coil is placed in close proximity to the first, and the mutual inductance of the coils is 130 µH. What is the peak emf (in V) in the second coil?

Answer 1

Answer:

The peak emf in second coil is 1.876 V

Explanation:

Given :

Inductance $L = 130 \times 10^{-6}$ H

The current $I(t) = 14 \sin(1.15\times 10^{3} t)$

We compare above equation with standard equation,

  $I(t) = I_{o} \sin (\omega t + \phi)$

From above equation we have,

  $\omega = 10^{3}$ and $\phi = 1.15$

Find the inductive resistance,

  $X_{L} = \omega L$

  $X_{L} = 10^{3} \times 130 \times 10^{-6}$

  $X_{L} = 0.134$

The peak emf in second coil is,

   $V = I_{o} X_{L}$

  $V = 14 \times 0.134$

  $V = 1.876$ V

Therefore, the peak emf in second coil is 1.876 V