Question

If a current flowing in a coil is reduced to half of its initial value, the relation between the new energy ($E_2$) and the original energy ($E_1$) stored in the coil will be

• A. $E_2 = \frac{E_1}{4}$
• B. $E_2 = \frac{E_1}{2}$
• C. $E_2 = E_1$
• D. $E_2 = 4E_1$

Hint:

Since energy is proportional to the square of the current ($E \propto I^2$), any factor changing the current will alter the energy by that factor squared. Cutting the current in half ($\frac{1}{2}$) means the energy scales down by a factor of $\left(\frac{1}{2}\right)^2 = \frac{1}{4}$ instantly!

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
An inductor coil stores magnetic potential energy when a current passes through it. We need to determine the relationship between the final stored energy $E_2$ and the initial energy $E_1$ when the circulating current is reduced to exactly half of its initial value.

Step 2: Key Formula or Approach:
The magnetic energy ($E$) stored inside an inductor loop carrying current $I$ is given by the formula: $$E = \frac{1}{2}LI^2$$ Where $L$ represents the self-inductance of the coil, which remains constant. This establishes a proportional relationship: $$E \propto I^2$$

Step 3: Detailed Explanation:
Let the initial energy with current $I_1$ be: $$E_1 = \frac{1}{2}LI_1^2$$ The current is halved, so the new current value is $I_2 = \frac{I_1}{2}$. Now write the expression for the new stored energy $E_2$: $$E_2 = \frac{1}{2}L(I_2)^2 = \frac{1}{2}L\left(\frac{I_1}{2}\right)^2$$ $$E_2 = \frac{1}{2}L\left(\frac{I_1^2}{4}\right) = \frac{1}{4}\left(\frac{1}{2}LI_1^2\right)$$ Substitute $E_1$ back into the brackets: $$E_2 = \frac{E_1}{4}$$

Step 4: Final Answer:
The new energy stored is $E_2 = \frac{E_1}{4}$, which matches option (A).