Question

A wire of length L meter carrying a current of I ampere is bent in the form of a circle, its magnetic moment will be
A. \[\dfrac{{iL}}{{4\pi }}\\ \]
B. \[\dfrac{{i{L^2}}}{{4\pi }}\\ \]
C. \[\dfrac{{{i^2}{L^2}}}{{4\pi }}\\ \]
D. \[\dfrac{{{i^2}L}}{{4\pi }}\\ \]

Answer 1

Hint:
The magnetic moment for a current-carrying loop is given as \[{\mu _m} = i.A\], where \[i\] is the current flowing in the loop and \[A\] is the area of the loop.
In this question, a wire of length L is given, which is further be bent to form a circular loop, so we can say the circumference of the loop will be equal to the length of the wire which is to be bent, then we find the radius of the loop from which we calculate the magnetic moment.

Complete step by step solution:
Given the length of the wire is L meter, which is then bent to form a circle, we know that the circumference of a circle is given as \[C = 2\pi r\]; hence we can say
\[L = 2\pi r - - (i)\]
This can be further written in the form of the radius of the circle as
\[r = \dfrac{L}{{2\pi }} - - (ii)\]

As we know, if the circulating current is \[i\]and the area enclosed by the current is \[A\] , then the magnetic moment \[{\mu _m}\] due to the circular coil will be
\[{\mu _m} = i.A - - (iii)\]
Now we can further write equation (iii) as
\[{\mu _m} = i.\pi .{r^2}\]
Where the value of radius \[r = \dfrac{L}{{2\pi }}\] from equation (ii), so we can write
\[{\mu _m} = i.\pi .{\left( {\dfrac{L}{{2\pi }}} \right)^2}\]
Now by further solving this, we get
\[
  {\mu _m} = i.\pi .\dfrac{{{L^2}}}{{4{\pi ^2}}} \\
   = \dfrac{{i.{L^2}}}{{4\pi }} \\
 \]
Hence the magnetic moment of the wire bent in the form of a circle is \[ = \dfrac{{i.{L^2}}}{{4\pi }}\]

Option (B) is correct

Note:
Students must know that whenever a material having some length to is bent to form a circular loop the circumference of the loop will be equal to the length of material.