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(A) The magnetic field at the centre of a circular loop of radius $r$ carrying current $i$ is $B_{loop} = \frac{\mu_0 i}{2r}$.The magnetic field at a

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$\frac{15}{29}$

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(A) The magnetic field at the centre of a circular loop of radius $r$ carrying current $i$ is $B_{loop} = \frac{\mu_0 i}{2r}$.The magnetic field at a distance $r$ from an infinitely long straight wire carrying current $i$ is $B_{wire} = \frac{\mu_0 i}{2\pi r}$.When the current in the loop is clockwise,the magnetic field due to the loop is directed into the plane of the paper,and the magnetic field due to the straight wire is directed out of the plane of the paper. Thus,$B_1 = \frac{\mu_0 i}{2r} - \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{2r} (1 - \frac{1}{\pi})$.When the current in the loop is anti-clockwise,the magnetic field due to the loop is directed out of the plane of the paper,and the magnetic field due to the straight wire is also directed out of the plane of the paper. Thus,$B_2 = \frac{\mu_0 i}{2r} + \frac{\mu_0 i}{2\pi r} = \frac{\mu_0 i}{2r} (1 + \frac{1}{\pi})$.Taking the ratio,$\frac{B_1}{B_2} = \frac{1 - \frac{1}{\pi}}{1 + \frac{1}{\pi}} = \frac{\pi - 1}{\pi + 1}$.Substituting $\pi \approx \frac{22}{7}$,we get $\frac{B_1}{B_2} = \frac{\frac{22}{7} - 1}{\frac{22}{7} + 1} = \frac{15}{29}$.

$A$ circular loop and an infinitely long straight conductor carry equal currents,as shown in the figure. The net magnetic field at the centre of the loop is $B_1$,when the current in the loop is clockwise and $B_2$ when the current in the loop is anti-clockwise. Then $\frac{B_1}{B_2}$ is

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