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Question

(A) The magnetic field on the axis of a circular loop of radius $R$ at a distance $d$ from its center is given by $B = \frac{\mu_0 I R^2}{2(R^2 + d^2)

Vedclass · Accepted Answer

$\frac{\mu_0 I}{16\sqrt{2}R} [(\sqrt{2} + 1)\hat{i} + \hat{j}]$

Vedclass · Answer

(A) The magnetic field on the axis of a circular loop of radius $R$ at a distance $d$ from its center is given by $B = \frac{\mu_0 I R^2}{2(R^2 + d^2)^{3/2}}$.Given $d = \sqrt{3}R$, we have $B = \frac{\mu_0 I R^2}{2(R^2 + 3R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(4R^2)^{3/2}} = \frac{\mu_0 I R^2}{2(8R^3)} = \frac{\mu_0 I}{16R}$.Let the first loop be in the $YZ$-plane with its axis along the $X$-axis. Its magnetic field at $P$ is $\vec{B}_1 = \frac{\mu_0 I}{16R} \hat{i}$.The second loop's axis is inclined at $45^\circ$ to the $X$-axis in the $XY$-plane. Its magnetic field vector $\vec{B}_2$ will be at $45^\circ$ to the $X$-axis: $\vec{B}_2 = \frac{\mu_0 I}{16R} (\cos 45^\circ \hat{i} + \sin 45^\circ \hat{j}) = \frac{\mu_0 I}{16R} (\frac{1}{\sqrt{2}} \hat{i} + \frac{1}{\sqrt{2}} \hat{j})$.The resultant field is $\vec{B} = \vec{B}_1 + \vec{B}_2 = \frac{\mu_0 I}{16R} [(1 + \frac{1}{\sqrt{2}})\hat{i} + \frac{1}{\sqrt{2}}\hat{j}] = \frac{\mu_0 I}{16R} [(\frac{\sqrt{2} + 1}{\sqrt{2}})\hat{i} + \frac{1}{\sqrt{2}}\hat{j}] = \frac{\mu_0 I}{16\sqrt{2}R} [(\sqrt{2} + 1)\hat{i} + \hat{j}]$.

Two identical circular current loops of radius $R$ carrying equal currents $I$ are placed such that their axes are inclined at $45^\circ$ to each other. The distance from the center of each loop to point $P$ is $\sqrt{3}R$. The resultant magnetic field at $P$ is:

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