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: Understanding the Question:
The question asks for the algebraic expression representing the total potential magnetic energy ($U$) stored in the surrounding magnetic field of an inductor coil.
The coil consists of $N$ turns, carries a steady excitation current $I$, and establishes a net magnetic flux linkage $\phi$ per turn.

Step 2: Key Formula or Approach:
1. The potential energy stored within the magnetic field of any inductor carrying current $I$ is given by:
$$U = \frac{1}{2} L I^2$$ 2. Self-inductance ($L$) is defined as the total magnetic flux linkage per unit of excitation current:
$$L = \frac{N\phi}{I}$$

Step 3: Detailed Explanation:
Let's substitute the definition of self-inductance ($L = \frac{N\phi}{I}$) directly into our standard magnetic energy equation:
$$U = \frac{1}{2} \left( \frac{N\phi}{I} \right) I^2$$ Simplify the expression by canceling one factor of current $I$ from the numerator and denominator:
$$U = \frac{1}{2} N \phi I = \frac{N\phi I}{2}$$ This matches the standard expression for stored inductive energy, analogous to the capacitive energy formula $U = \frac{1}{2}QV$.

Step 4: Final Answer:
The magnetic energy stored in the medium surrounding the coil is $\frac{N\phi I}{2}$, which corresponds to option (B).