Question

At the instant a current of 0.40 A is flowing through a coil of wire, the energy stored in its magnetic field is 9.5 ✕ 10−3 J. What is the self-inductance of the coil (in H)?

Answer 1

The self-inductance of the coil with a current of 0.40 A is flowing through a coil of wire, the energy stored in its magnetic field is 9.5 *10^−3 J is 0.12 H.

To find the answer, we have to know about the energy stored in the magnetic field inductor.

How to find the self inductance of the coil?

• We have expression of energy stored in the magnetic field inductor as,

                      $E=\frac{LI^2}{2}$

• Thus, by rearranging it, we get the self inductance as,

                       $L=\frac{2E}{I^2} =\frac{2*9.5*10^{-3}}{0.40^2} \\L=0.12H$

Thus, we can conclude that, the self-inductance of the coil with a current of 0.40 A is flowing through a coil of wire, the energy stored in its magnetic field is 9.5 *10^−3 J is 0.12 H.

Learn more about the energy stored in the magnetic field inductor here: https://brainly.com/question/17176606

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