Question

A circular coil is made from a wire of length 2 m. Its radius is $\dfrac{4}{\pi }cm$. When a current of 1 A passes through it, its magnetic moment is
$\begin{align}
  & \text{A}\text{. }\pi A{{m}^{2}} \\
 & \text{B}\text{. }\dfrac{4}{100\pi }A{{m}^{2}} \\
 & \text{C}\text{. }\dfrac{16}{\pi }A{{m}^{2}} \\
 & \text{D}\text{. }\dfrac{1}{\pi }A{{m}^{2}} \\
\end{align}$

Answer 1

Hint: We are given the length of the wire used to make a circular coil and also its radius. It is said that a 1 A current is passed through the coil and we are asked to calculate the magnetic moment. We know that by turning the wire ‘n’ times we will the circular coil of the given radius. Using this relation we will get the number of turns and by substituting all the known values in the equation for magnetic moment we will get the solution.

Formula used: $L=n\times 2\pi r$
$m=nIA$

Complete step by step answer:
In the question we have a circular coil made from a wire of length 2 m and the radius of the circular coil is $\dfrac{4}{\pi }cm$.
Let ‘n’ be the number of turns, ‘r’ be the radius and ‘L’ be the length of the wire. Then we have,
$L=n\times 2\pi r$
From the question we have,
$L=2m$
$r=\dfrac{4}{\pi }cm=\dfrac{4}{100\pi }m$
Therefore we can write,
$\Rightarrow 2=n\left( 2\pi \dfrac{4}{100\pi } \right)$
By solving this equation we get the value of number of turns as,
$\Rightarrow n=\left( \dfrac{2}{2\pi \times \dfrac{4}{100\pi }} \right)$
$\Rightarrow n=\left( \dfrac{100}{4} \right)=25$
Therefore we get the number of turns of the coil as 25.
In the question it is said that a current of 1 A is passed through the coil and we are asked to find the magnetic moment in the coil.
We know that the magnetic moment is given by the equation,
$m=nIA$, where ‘n’ is the number of turns, ‘I’ is the current passed and ‘A’ is the area of the cross section.
Here for the circular coil we have area of the cross section as,
$A=\pi {{r}^{2}}$
Therefore we get the magnetic moment as,
$\Rightarrow m=25\times 1\times \pi \times {{\left( \dfrac{4}{100\pi } \right)}^{2}}$
By solving this we get,
$\Rightarrow m=25\times 1\times \pi \times \dfrac{16}{10000{{\pi }^{2}}}$
$\Rightarrow m=25\times \dfrac{16}{10000\pi }$
$\Rightarrow m=\dfrac{400}{10000\pi }$
$\Rightarrow m=\dfrac{4}{100\pi }A{{m}^{2}}$
Therefore the magnetic moment in the coil when we pass 1 a current through it is $\dfrac{4}{100\pi }A{{m}^{2}}$.

So, the correct answer is “Option B”.

Note: A magnetic dipole is formed when we arrange two unlike magnetic poles which are of equal pole strength at a distance from each other.
Magnetic moment is simply a vector quantity (a quantity that has both direction and magnitude) that gives us the tendency of an object in a magnetic field to interact with an external magnetic field.