Question

If current of 4 A produces magnetic flux of \(3\times 10^{-3}\) Wb through a coil of 400 turns, the energy stored in the coil will be

• A. \(1.2\) J
• B. \(2.4\) J
• C. \(24\) J
• D. \(240\) J

Hint:

Find L = N phi / I, then U = (1/2) L I^2.

Answer 1

The Correct Option is B

Step 1: Use a shortcut:
$U = \tfrac12 N\phi I$, because $LI = N\phi$.

Step 2: Substitute:
$U = \tfrac12\times400\times3\times10^{-3}\times4 = \tfrac12\times4.8 = 2.4$ J.

Step 3: Cross-check:
$L = N\phi/I = 0.3$ H and $\tfrac12LI^2 = \tfrac12\times0.3\times16 = 2.4$ J. Both routes agree.

Step 4: Cross-check with flux linkage:
The flux linkage is $N\phi = 400\times3\times10^{-3} = 1.2$ Wb. Then $L = 1.2/4 = 0.3$ H, and the energy $\frac12\times1.2\times4 = 2.4$ J. All routes give the same energy.

Final Answer:
Option (B). \[ \boxed{2.4 \text{ J (B)}} \]