Question

A coil having 0.64 mH inductance and 0.8 \(\Omega\) resistance connected to a battery of 12 V. The energy stored in the magnetic field created by that coil is:

• A. \(32 \times 10^{-3} \, {J}\)
• B. \(81 \times 10^{-3} \, {J}\)
• C. \(64 \times 10^{-3} \, {J}\)
• D. \(72 \times 10^{-3} \, {J}\)

Hint:

To find the energy stored in the magnetic field of an inductor, use the formula \(E = \frac{1}{2} L I^2\), where \(L\) is the inductance and \(I\) is the current. Make sure to calculate the current using Ohm’s law if the voltage and resistance are given.

Answer 1

The Correct Option is D

We are given the inductance \(L = 0.64 \, {mH} = 0.64 \times 10^{-3} \, {H}\), the resistance \(R = 0.8 \, \Omega\), and the voltage \(V = 12 \, {V}\). 
To find the energy stored in the magnetic field, we use the formula: \[ E = \frac{1}{2} L I^2 \] First, calculate the current \(I\) using Ohm’s law: \[ I = \frac{V}{R} = \frac{12}{0.8} = 15 \, {A} \] Now, substitute the values into the energy formula: \[ E = \frac{1}{2} \times 0.64 \times 10^{-3} \times (15)^2 \] \[ E = \frac{1}{2} \times 0.64 \times 10^{-3} \times 225 = 72 \times 10^{-3} \, {J} \] Thus, the energy stored in the magnetic field is \(72 \times 10^{-3} \, {J}\).