MHT CET · Physics · Electromagnetic Induction
When a current \(I\) flows through a coil, it stores the energy \(E_1\) in it. Now the current is reduced to \(\frac{l}{2}\), the energy stored in the coil becomes \(E_2\). The change in the energy is
- A \(\frac{E_1}{4}\)
- B \(\frac{3 E_1}{4}\)
- C \(\frac{4 E_1}{3}\)
- D \(\frac{E_1}{2}\)
Answer & Solution
Correct Answer
(B) \(\frac{3 E_1}{4}\)
Step-by-step Solution
Detailed explanation
Energy stored in the coil \(E \propto I^2\)
\(\therefore E_1 \propto I^2 \text { and } E_2 \propto\left(\frac{I}{2}\right)^2\)
On taking the ratio:
\(E_1=4 E_2\)
The change in energy is \(E_1-E_2=E_1-\frac{E_1}{4}=\frac{3 E_1}{4}\)
\(\therefore E_1 \propto I^2 \text { and } E_2 \propto\left(\frac{I}{2}\right)^2\)
On taking the ratio:
\(E_1=4 E_2\)
The change in energy is \(E_1-E_2=E_1-\frac{E_1}{4}=\frac{3 E_1}{4}\)
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