In the diagram,I_{1} and I_{2} are the streng | vedclass

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

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$\frac{1}{2 \pi}$

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(D) The magnetic field at the centre $O$ due to a circular loop of radius $R$ carrying current $I_{1}$ is given by: $B_{1} = \frac{\mu_{0} I_{1}}{2 R}$.The magnetic field at the centre $O$ due to a long straight conductor carrying current $I_{2}$ at a perpendicular distance $d = OA + AB = R + R = 2R$ is given by: $B_{2} = \frac{\mu_{0} I_{2}}{2 \pi d} = \frac{\mu_{0} I_{2}}{2 \pi (2R)} = \frac{\mu_{0} I_{2}}{4 \pi R}$.Given that the net magnetic field at $O$ is zero,the magnitudes of the magnetic fields produced by the loop and the straight conductor must be equal: $B_{1} = B_{2}$.Substituting the expressions: $\frac{\mu_{0} I_{1}}{2 R} = \frac{\mu_{0} I_{2}}{4 \pi R}$.Simplifying the equation: $\frac{I_{1}}{2} = \frac{I_{2}}{4 \pi}$.Therefore,the ratio of the currents is: $\frac{I_{1}}{I_{2}} = \frac{2}{4 \pi} = \frac{1}{2 \pi}$.

In the diagram,$I_{1}$ and $I_{2}$ are the strengths of the currents in the loop and straight conductor,respectively. Given $OA = AB = R$. The net magnetic field at the centre $O$ is zero. Then the ratio of the currents in the loop and the straight conductor is:

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