Two wires $A$ and $B$ are of lengths $40 \mathrm{~cm}$ and $30 \mathrm{~cm}$. $A$ is bent into a circle of…
Two wires $A$ and $B$ are of lengths $40 \mathrm{~cm}$ and $30 \mathrm{~cm}$. $A$ is bent into a circle of radius $r$ and $B$ into an arc of radius $r$. A current $i_1$ is passed through $A$ and $i_2$ through $B$. To have the same magnetic inductions at the centre, the ratio of $i_1: i_2$ is
$3: 4$
$3: 5$
$2: 3$
$4: 3$
Solution
For wire $A$,
$B_1=\frac{\mu_0 \dot{l}_1}{2 r}$
where
$r=\frac{40}{2 \pi}$
For wire $B$,
$\begin{aligned} \text { Circumference } & =\text { length } \\ n \pi r & =30\end{aligned}$
or $\quad n \pi=\frac{30}{r}=\frac{30}{40 / 2 \pi}=\frac{3}{2} \pi$
$\theta=n \pi=\frac{3}{2} \pi$
$\therefore \quad B_2=\frac{\mu_0}{4 \pi}\left(\frac{i_2}{r}\right) \theta$
But $\quad B_1=B_2$
$\begin{array}{rlrl} & \text { or } & \frac{\mu_0 i_1}{2 r} & =\frac{\mu_0}{4 \pi}\left(\frac{i_2}{r}\right) \theta \\ \text { or } & \frac{i_1}{i_2} & =\frac{3}{4}\end{array}$