Question

A metallic wire of length $10\text{cm}$ , when made into a circular loop of radius $R$ and a current of $i\text{A}$ flows through it gives same magnetic moment value as when it is made into a square of side $a$. The relation between the current passing through the square loop $i^{\prime }$ to the current passing through the circular loop are given by $i=x{i}^{\prime }$. Find $x$.

• A. $\frac{\pi }{4}$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{5}$
• D. $\frac{\pi }{16}$

Answer 1

The correct option is A $\frac{\pi }{4}$

Given that, Magnetic moment for both circular loop and square loop are same.

Also, $L=10\text{cm}$

$\therefore I_1{A}_1=I_2{A}_2$

$\Rightarrow i\left(\pi R^2\right)=i^{\prime }(a{)}^2$

But, $L=2\pi R$ and also $L=4a$

Thus, $R=\frac{5}{\pi }\text{cm}$ and $a=2.5\text{cm}$

Substituting these in the above equation we get,

$i=i^{\prime }\times \left(\frac{(2.5{)}^2}{\pi \times (5/\pi {)}^2}\right)=\frac{\pi }{4}{i}^{\prime }$

comparing this with the data given in the question we get, $x=\frac{\pi }{4}$

Hence, option (a) is the correct answer.