Question

The magnetic potential energy stored in a certain inductor is $25\ \text{mJ}$, when the current in the inductor is $50\ \text{mA}$. This inductor is of inductance

• A. $2.00\ \text{H}$
• B. $0.20\ \text{H}$
• C. $200\ \text{H}$
• D. $20\ \text{H}$

Hint:

Always ensure units are converted to standard SI units (Joules for energy, Amperes for current) before substituting into the formula to avoid power-of-ten errors.

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
We are given the magnetic potential energy ($U$) stored in an inductor and the current ($I$) flowing through it. We need to calculate the self-inductance ($L$) of the inductor.

Step 2: Key Formula or Approach:
The magnetic potential energy $U$ stored in an inductor is given by the formula $U = \frac{1}{2} L I^2$. Rearranging to solve for inductance: $L = \frac{2U}{I^2}$.

Step 3: Detailed Explanation:
Given values (converted to SI units):
$U = 25\ \text{mJ} = 25 \times 10^{-3}\ \text{J}$
$I = 50\ \text{mA} = 50 \times 10^{-3}\ \text{A} = 5 \times 10^{-2}\ \text{A}$
Now, substitute these into the rearranged formula:
$L = \frac{2 \times (25 \times 10^{-3})}{(5 \times 10^{-2})^2}$
$L = \frac{50 \times 10^{-3}}{25 \times 10^{-4}}$
$L = \frac{50}{25} \times \frac{10^{-3}}{10^{-4}} = 2 \times 10^1 = 20\ \text{H}$.

Step 4: Final Answer:
The inductance of the inductor is $20\ \text{H}$, which corresponds to option (D).