A battery is connected between two points A and $B$ on the circumference of a uniform conducting rin

Question

A battery is connected between two points A and $B$ on the circumference of a uniform conducting ring of radius $r$ and resistance $R$. One of the arcs $A B$ of the ring subtends on angle $	heta$ at the center as shown in figure. The value of the magnetic field at the center due to current in the ring is\

Professor PC Thomas Classes · Answer

Let $I_{1}$ and $I_{2} b$ e the currents in two parts of the ring and $\mathrm{R}_{1}$ and $\mathrm{R}_{2}$ be the resistances of two parts of ring. Then $\mathrm{I}_{1} \mathrm{R}_{1}=\mathrm{I}_{2} \mathrm{R}_{2}, \mathrm{R} \alpha \ell$\ $	herefore \mathrm{I}_{1} \ell_{1}=\mathrm{I}_{2} \ell_{2}$\ $\mathrm{B}_{1}=\frac{\mu_{0} \mathrm{I}_{1}}{4 \pi \mathrm{r}} 	heta_{1}=\frac{\mu_{0} \mathrm{I}_{1} \ell_{1}}{4 \pi \mathrm{r}}$\ $\mathrm{B}_{2}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{2} 	heta_{2}}{\mathrm{r}}=\frac{\mu_{0}}{4 \pi} \frac{\mathrm{I}_{2} \ell_{2}}{\mathrm{r}}$\ since $I_{1} \ell_{1}=I_{2} \ell_{2}, B_{1}=B_{2}$\