A current I enters a circular coil of radius | vedclass

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Question

Let $l_{1}, l_{2}$ be the length of the two part $\mathrm{ABC}$ and $ADC$ of the conductor and $\rho$ be resistance per unit lenght of the conductor:

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Let $l_{1}, l_{2}$ be the length of the two part $\mathrm{ABC}$ and $ADC$ of the conductor and $ho$ be resistance per unit lenght of the conductor:

The resistance of the part $\mathrm{ABC}$ will be: $\mathrm{R}_{1}=ho l_{1}$

The resistance of the part $\mathrm{ADC}$ will be: $\mathrm{R}_{2}=ho l_{2}$

Potential difference across

$\mathrm{AC}=\mathrm{I}_{1} \mathrm{R}_{1}=\mathrm{I}_{2} \mathrm{R}_{2}$

or $I_{1} ho l_{1}=I_{2} ho l_{2}$

or $\mathrm{I}_{1} l_{1}=\mathrm{I}_{2} l_{2}.........(i)$

Magnetic field at $O$ due to a current $I_{1}$ flowing in $ABC$ is given by:

$\mathrm{B}_{1}=\frac{\mu_{0} I_{1} 	heta_{1}}{4 \pi R}=\frac{\mu_{0} I_{1} l_{1}}{4 \pi R^{2}} \otimes \quad\left(	ext { As } l_{1}=	heta_{1} Right)$

Magnetic field at $O$ due to current $I_2$, flowing in $\mathrm{ADC}$ is given by :-

$\mathrm{B}_{2}=\frac{\mu_{0} I_{2} 	heta_{2}}{4 \pi R}=\frac{\mu_{0} I_{2} l_{2}}{4 \pi R^{2}} \odot \quad\left(	ext { As } l_{2}=	heta_{2} Right)$

or $\left.\quad \mathrm{B}_{2}=\frac{\mu_{0} \mathrm{I}_{1} l_{1}}{4 \pi \mathrm{R}^{2}} \odot \quad 	ext { [Using eqn. (i) }ight]$

The resultant magnetic field induction at $\mathrm{O}$ is, $\mathrm{B}=\mathrm{B}_{2}-\mathrm{B}_{1}=0$

$\left(\mathrm{As} \,\,\,\mathrm{B}_{2}=\mathrm{B}_{1}ight)$

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 at the centre of the coil is

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