Question

A closely wound solenoid of 800 turns and area of cross section $2.5 \times 10^{-4}$ m$^2$ can carry a current of 3.0 A. The magnetic moment associated with it is ______.

• A. 60 JT$^{-1}$
• B. 0.60 JT$^{-1}$
• C. 6 JT$^{-1}$
• D. 0.06 JT$^{-1}$

Hint:

The unit $\text{A}\cdot\text{m}^2$ is equivalent to $\text{J/T}$ (Joules per Tesla) because potential energy $U = -\vec{M} \cdot \vec{B}$.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
The magnetic moment ($M$) of a solenoid or coil is the product of the number of turns, the current, and the area of the loop.
Step 2: Calculation:
Formula: $M = NIA$ Given: $N = 800$, $I = 3.0\text{ A}$, $A = 2.5 \times 10^{-4}\text{ m}^2$. $$M = 800 \times 3.0 \times 2.5 \times 10^{-4}$$ $$M = 2400 \times 2.5 \times 10^{-4}$$ $$M = 6000 \times 10^{-4} = 0.60\text{ A}\cdot\text{m}^2 \text{ (or JT}^{-1})$$
Step 3: Final Answer:
The magnetic moment is 0.60 JT$^{-1}$.