Question

If an inductor coil of self-inductance 2 H stores 25 J of magnetic energy, then the current I passing through it is:

• A. 25 A
• B. 10 A
• C. 15 A
• D. 2 A
• E. 5 A

Hint:

This is the magnetic equivalent of capacitive energy $1/2 CV^2$.

Answer 1

Answer: (E)

Step 1: Understanding the Concept:
An inductor stores energy in the magnetic field created when current flows through it. This problem requires using the formula that relates the stored magnetic energy to the inductance and the current.
Step 2: Key Formula or Approach:
The energy (U) stored in an inductor is given by: \[ U = \frac{1}{2} L I^2 \] where L is the self-inductance and I is the current flowing through it. We need to rearrange this formula to solve for the current, I. \[ I^2 = \frac{2U}{L} \implies I = \sqrt{\frac{2U}{L}} \] Step 3: Detailed Explanation:
We are given: - Self-inductance, L = 2 H - Stored energy, U = 25 J Substitute these values into the rearranged formula: \[ I = \sqrt{\frac{2 \times 25}{2}} \] Cancel the 2s: \[ I = \sqrt{25} \] \[ I = 5 \text{ A} \] Step 4: Final Answer:
The current passing through the inductor is 5 A.