Two circular loops L_1 and L_2 of wire carryi | vedclass

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(D) The magnetic field at a point on the axis of a circular loop of radius $R_1$ carrying current $i$ at a distance $x$ from its centre is given by $B

Vedclass · Accepted Answer

$R_2 = 8 R_1$

Vedclass · Answer

(D) The magnetic field at a point on the axis of a circular loop of radius $R_1$ carrying current $i$ at a distance $x$ from its centre is given by $B = \frac{\mu_0 i R_1^2}{2(R_1^2 + x^2)^{3/2}}$.Here,the distance between the centres is $x = \sqrt{3} R_1$.So,the magnetic field $B_1$ produced by loop $L_1$ at the centre of loop $L_2$ is:$B_1 = \frac{\mu_0 i R_1^2}{2(R_1^2 + (\sqrt{3} R_1)^2)^{3/2}} = \frac{\mu_0 i R_1^2}{2(R_1^2 + 3 R_1^2)^{3/2}} = \frac{\mu_0 i R_1^2}{2(4 R_1^2)^{3/2}} = \frac{\mu_0 i R_1^2}{2(8 R_1^3)} = \frac{\mu_0 i}{16 R_1}$.The magnetic field $B_2$ at the centre of loop $L_2$ due to its own current $i$ is $B_2 = \frac{\mu_0 i}{2 R_2}$.For the net magnetic field at the centre of $L_2$ to be zero,the magnitudes of $B_1$ and $B_2$ must be equal because the currents are opposite:$B_1 = B_2 \implies \frac{\mu_0 i}{16 R_1} = \frac{\mu_0 i}{2 R_2}$.Solving for $R_2$,we get $R_2 = 8 R_1$.

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