In the given diagram,two current-carrying cir | vedclass

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(B) The magnetic field at the center of a circular loop of radius $r$ carrying current $I$ is $B = \frac{\mu_0 I}{2r}$.For the loop in the $YZ-$ plane

Vedclass · Accepted Answer

$\sin^{-1}\left(\frac{1}{\sqrt{5}}\right)$

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(B) The magnetic field at the center of a circular loop of radius $r$ carrying current $I$ is $B = \frac{\mu_0 I}{2r}$.For the loop in the $YZ-$ plane (radius $R$),the magnetic field is directed along the $X-$ axis: $B_x = \frac{\mu_0 I}{2R}$.For the loop in the $XZ-$ plane (radius $2R$),the magnetic field is directed along the $Y-$ axis: $B_y = \frac{\mu_0 I}{2(2R)} = \frac{\mu_0 I}{4R}$.The resultant magnetic field vector is $\vec{B} = B_x \hat{i} + B_y \hat{j}$.The angle $	heta$ with the $X-$ axis is given by $	an 	heta = \frac{B_y}{B_x} = \frac{\mu_0 I / 4R}{\mu_0 I / 2R} = \frac{2R}{4R} = \frac{1}{2}$.Since $	an 	heta = \frac{1}{2}$,we have a right triangle with opposite side $1$ and adjacent side $2$. The hypotenuse is $\sqrt{1^2 + 2^2} = \sqrt{5}$.Therefore,$\sin 	heta = \frac{	ext{opposite}}{	ext{hypotenuse}} = \frac{1}{\sqrt{5}}$,which implies $	heta = \sin^{-1}\left(\frac{1}{\sqrt{5}}ight)$.

In the given diagram,two current-carrying circular loops of radius $R$ and $2R$ are arranged in the $YZ-$ plane and $XZ-$ plane respectively. The common center of both is at the origin $O$. What will be the angle of the resultant magnetic field from the $X-$ axis?

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