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. $\frac{{\mu }_{0}i{l}_1{l}_2}{2R{\left(l_1+l_2\right)}^2}$
• B. $\frac{{\mu }_0{i}_1{l}_2}{4\pi R^2\left(l_1+l_2\right)}$
• C. $\frac{{\mu }_{0}i\left(l_1+l_2\right)}{2\pi R{l}_1{l}_2}$
• D. Zero

Answer 1

Answer: (B)

Let current in part $ABC$ is $i_1$
and in part $ADC$ is $i_2$
$i=\frac{i{l}_2}{l_1+l_2}$
(As $ABC$ and $ADC$ part are in parallel connection)
and subtended by $ABC$ at centre $O$ will be $=\left(\frac{2\pi }{l_1+l_2}\right)\left(l_1\right)$
so using $B=\frac{{\mu }_{0}i}{2a}\left(\frac{\theta }{2\pi }\right)$
$B=\frac{{\mu }_0}{2R}\left(\frac{i{l}_2}{l_1+l_2}\right)\frac{2\pi }{\left(l_1+l_2\right)}\frac{\left(l_1\right)}{2\pi }$