Question

A uniform circular wire loop is connected to the terminals of a battery. The magnetic field induction at the centre due to $ABC$ portion of the wire will be (length of $ABC = l _{1}$, length of $ADC = l _{2}$ )

• A. $\frac{\mu_{0} il _{1} l _{2}}{2 R \left( l _{1}+ l _{2}\right)^{2}}$
• B. $\frac{\mu_{0} i _{1} l _{2}}{4 \pi R ^{2}\left(l_{1}+ l _{2}\right)}$
• C. $\frac{\mu_{0} i \left(l_{1}+ l _{2}\right)}{2 \pi R \quad l _{1} l _{2}}$
• D. Zero

Answer 1

Answer: (B)

Let current in part $A B C$ is $i_{1}$
and in part $A D C$ is $i_{2}$
$i=\frac{i l_{2}}{l_{1}+l_{2}} $
(As $A B C$ and $A D C$ part are in parallel connection)
and subtended by $A B C$ at centre $O$ will be $=\left(\frac{2 \pi}{l_{1}+l_{2}}\right)\left(l_{1}\right)$
so using $B=\frac{\mu_{0} i}{2 a}\left(\frac{\theta}{2 \pi}\right)$
$B=\frac{\mu_{0}}{2 R}\left(\frac{i l_{2}}{l_{1}+l_{2}}\right) \frac{2 \pi}{\left(l_{1}+l_{2}\right)} \frac{\left(l_{1}\right)}{2 \pi}$