JEE Mains · Physics · STD 12 -6. Electromagnetic induction
A coil has an inductance of \(2 H\) and resistance of \(4\,\Omega\). A \(10\,V\) is applied across the coil. The energy stored in the magnetic field after the current has built up to its equilibrium value will be \(..........\times 10^{-2} J\)
- A \(625\)
- B \(624\)
- C \(623\)
- D \(621\)
Answer & Solution
Correct Answer
(A) \(625\)
Step-by-step Solution
Detailed explanation
\(I =\frac{ V }{ R }=\frac{5}{2}\,A\) \(E =\frac{1}{2} LI ^2=\frac{1}{2} \times 2 \times\left(\frac{5}{2}\right)^2\) \(E =625 \times 10^{-2}\,J\)
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