Question

A wire of length $L$ metre carrying a current of $I$ ampere is bent in the form of a circle. Its magnetic moment is,

• A. $I \,L^{2} /4 \,A\, m^{2}$
• B. $1\,\pi\,L^{2}/4\,A\,m^{2}$
• C. $2\,I\,L^{2}/\pi\, A\,m^{2}$
• D. $I\,L^{2}/4\pi\,A\,m^{2}$

Answer 1

The Correct Option is D

To determine the magnetic moment of a wire bent into a circle, follow these steps:

1. Understanding the Formula:

The magnetic moment \(M\) of a circular loop carrying a current \(I\) is given by:

\(M = I \times A\)

where \(A\) is the area of the circular loop.

1. Calculate the Radius of the Circle:

Since the wire of length \(L\) is bent into a circle, the circumference of the circle is equal to \(L\):

\(2\pi r = L\)

Solving for \(r\) (the radius of the circle):

\(r = \frac{L}{2\pi}\)

1. Calculate the Area of the Circle:

The area \(A\) of the circle is given by:

\(A = \pi r^2\)

Substituting for \(r\):

\(A = \pi \left( \frac{L}{2\pi} \right)^2 = \frac{\pi L^2}{4\pi^2} = \frac{L^2}{4\pi}\)

1. Calculate the Magnetic Moment:

Using the magnetic moment formula \(M = I \times A\):

\(M = I \times \frac{L^2}{4\pi} = \frac{I L^2}{4\pi}\)

1. Conclusion:

The magnetic moment of the wire when bent into the form of a circle is \(\frac{I L^2}{4\pi} \, A \, m^2\), which matches the correct answer option given: \(\frac{I L^2}{4\pi} \, A \, m^2\).