Question

For a given length \(L\) of a wire carrying a current \(I\), how many circular turns will produce the maximum magnetic moment?

• A. \(1\)
• B. \(L\)
• C. \(LI\)
• D. \(L+I\)

Hint:

For fixed wire length, \[ M=NIA = \frac{IL^2}{4\pi N}. \] Thus magnetic moment decreases as the number of turns increases.

Answer 1

The Correct Option is A

Concept: Magnetic moment of a current loop is \[ M=NIA \] where \(N\) is the number of turns and \(A\) is the area of each turn.

Step 1: Express area in terms of total wire length. If the wire makes \(N\) circular turns, \[ L=N(2\pi r) \] \[ r=\frac{L}{2\pi N}. \] Area of each turn: \[ A=\pi r^2 \] \[ = \pi \left( \frac{L}{2\pi N} \right)^2 \] \[ = \frac{L^2}{4\pi N^2}. \]

Step 2: Write magnetic moment. \[ M = NIA \] \[ = N I \left( \frac{L^2}{4\pi N^2} \right) \] \[ = \frac{IL^2}{4\pi N}. \]

Step 3: Determine the value of \(N\). \[ M\propto\frac1N. \] Hence magnetic moment is maximum when \(N\) is minimum. The smallest possible number of turns is \[ N=1. \] \[\begin{aligned} \boxed{1} \end{aligned}\] Hence, option \(\mathbf{(A)}\) is correct.