Question

A wire of length $L$ is bent into a circular loop and carries current $I$. What is the magnetic moment of the loop?

Answer 1

The magnetic moment of the loop is $\frac{IL^2}{4\pi}$.

Answer 2

The magnetic moment of a current loop is given by the product of the current flowing through it and the area enclosed by the loop.

1. Relate the length of the wire to the radius of the circular loop: When a wire of length $L$ is bent into a circular loop, its length becomes the circumference of the circle. Let $R$ be the radius of the circular loop. The circumference of the loop is $2\pi R$. Therefore, we have: $L = 2\pi R$ From this, we can express the radius $R$ in terms of $L$: $R = \frac{L}{2\pi}$

2. Calculate the area of the circular loop: The area $A$ of a circular loop with radius $R$ is given by: $A = \pi R^2$ Substitute the expression for $R$ from step 1 into this formula: $A = \pi \left(\frac{L}{2\pi}\right)^2$ $A = \pi \frac{L^2}{4\pi^2}$ $A = \frac{L^2}{4\pi}$

3. Calculate the magnetic moment of the loop: The magnetic moment $M$ of a current loop is given by the formula: $M = IA$ Where $I$ is the current flowing through the loop and $A$ is the area of the loop. Substitute the expression for $A$ from step 2 into this formula: $M = I \left(\frac{L^2}{4\pi}\right)$ $M = \frac{IL^2}{4\pi}$