Question

If the current of $I\text{ A}$ gives rise to a magnetic flux $\phi$ through a coil having $N$ turns, then magnetic energy stored in the medium surrounding the coil is

• A. $\frac{N\phi I}{4}$
• B. $\frac{N\phi I}{2}$
• C. $NI^2\phi$
• D. $N\phi^2 I$

Hint:

Think of the electrical-magnetic analogy to remember this formula instantly! The energy stored in a capacitor is $\frac{1}{2}QV$, where $Q$ is the total charge carrier value and $V$ is potential. In magnetism, the total charge-equivalent parameter is the total flux linkage ($N\phi$), and the potential-equivalent parameter is the current ($I$). Replacing these variables yields $\frac{1}{2}(N\phi)I$.

Answer 1

The Correct Option is B

Step 1: Energy in an inductor.
The magnetic energy stored in a coil of self-inductance $L$ carrying current $I$ is \[ U = \tfrac{1}{2}LI^2. \]
Step 2: Define self-inductance through flux linkage.
The total flux linkage is $N\phi$, and inductance is flux linkage per unit current: \[ L = \frac{N\phi}{I}. \]
Step 3: Substitute $L$ into the energy.
\[ U = \tfrac{1}{2}\left(\frac{N\phi}{I}\right)I^2. \]
Step 4: Cancel one current factor.
\[ U = \tfrac{1}{2}N\phi I. \]
Step 5: Note the analogy.
This mirrors the capacitor result $U = \tfrac{1}{2}QV$, with flux linkage playing the role of charge and current the role of voltage.
Step 6: State the answer.
\[ U = \frac{N\phi I}{2}, \] which is option (B). \[ \boxed{U = \frac{N\phi I}{2}} \]