A circular coil of radius R is carrying a cur | vedclass

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(C) The magnetic field at the centre of the circular coil due to current $I_{1}$ is given by: $B_{1} = \frac{\mu_{0} I_{1}}{2 R}$ The magnetic field a

Vedclass · Accepted Answer

$\frac{R}{\pi}\left(\frac{I_{2}}{I_{1}}\right)$

Vedclass · Answer

(C) The magnetic field at the centre of the circular coil due to current $I_{1}$ is given by: $B_{1} = \frac{\mu_{0} I_{1}}{2 R}$ The magnetic field at the centre of the coil due to the long straight wire carrying current $I_{2}$ at a distance $d$ is given by: $B_{2} = \frac{\mu_{0} I_{2}}{2 \pi d}$ For the net magnetic field at the centre to be zero,the magnitudes of these fields must be equal and their directions must be opposite: $B_{1} = B_{2}$ Substituting the expressions: $\frac{\mu_{0} I_{1}}{2 R} = \frac{\mu_{0} I_{2}}{2 \pi d}$ Solving for $d$: $d = \frac{\mu_{0} I_{2}}{2 \pi} 	imes \frac{2 R}{\mu_{0} I_{1}}$ $d = \frac{I_{2} R}{\pi I_{1}}$ Thus,the correct option is $C$.

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