A current I enters a circular coil of radius $R$, branches into two parts and then recombines as sho

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A current I enters a circular coil of radius $R$, branches into two parts and then recombines as shown in the circuit diagram.\ The resultant magnetic field is\

Professor PC Thomas Classes · Answer

Let $\ell_{1}, \ell_{2}$ be the length of the two parts ABC and ADC of the conductor and $ho$ be resistance per unit length of the conductor.\ The resistance of the part ABC will be $R_{1}=ho \ell_{1}$ $\mathrm{I}_{1} ho \ell_{1}=\mathrm{I}_{2} ho \ell_{2}$ or $\mathrm{I}_{1} \ell_{1}=\mathrm{I}_{2} \ell_{2} \quad \ldots$ (i)\ Magnetic field at O due to a current $\mathrm{I}_{1}$ flowing in ABC is given by\ $B_{1}=\frac{\mu_{0} I_{1} 	heta_{1}}{4 \pi R}=\frac{\mu_{0} I_{1} \ell_{1}}{4 \pi R^{2}} \otimes\left(ight.$ As $\left.\ell_{1}=	heta_{1} Right)$ Magnetic field at O due to current $\mathrm{I}_{2}$ flowing in ADC is given by\ $B_{2}=\frac{\mu_{0} I_{2} 	heta_{2}}{4 \pi R}=\frac{\mu_{0}}{4 \pi} \frac{I_{2} \ell_{2}}{R^{2}} \odot \quad\left(ight.$ As $\left.\ell_{2}=	heta_{2} Right)$\ $\mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{1} \ell_{1}}{4 \pi \mathrm{R}^{2}} \odot[$ using (i) $]$\ The resultant magnetic field induction at $O$ is\ $\mathrm{B}=\mathrm{B}_{2}-\mathrm{B}_{1}=0\left(\mathrm{As} \mathrm{B}_{2}=\mathrm{B}_{1}ight)$\