Two concentric circular loops of radii r_{1} | vedclass

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(B) The magnetic moment of a current loop is given by $\vec{M} = I \vec{A}$,where $\vec{A}$ is the area vector directed according to the right-hand ru

Vedclass · Accepted Answer

$-\frac{7}{2} \hat{k} \, A \cdot m^{2}$

Vedclass · Answer

(B) The magnetic moment of a current loop is given by $\vec{M} = I \vec{A}$,where $\vec{A}$ is the area vector directed according to the right-hand rule.For the inner loop (radius $r_{1} = 0.3 \, m$),the current flows counter-clockwise,so $\vec{M}_{1} = I \pi r_{1}^{2} \hat{k} = 7 	imes \pi 	imes (0.3)^{2} \hat{k} = 0.63 \pi \hat{k} \, A \cdot m^{2}$.For the outer loop (radius $r_{2} = 0.5 \, m$),the current flows clockwise,so $\vec{M}_{2} = -I \pi r_{2}^{2} \hat{k} = -7 	imes \pi 	imes (0.5)^{2} \hat{k} = -1.75 \pi \hat{k} \, A \cdot m^{2}$.The net magnetic moment is $\vec{M}_{net} = \vec{M}_{1} + \vec{M}_{2} = (0.63 \pi - 1.75 \pi) \hat{k} = -1.12 \pi \hat{k} \, A \cdot m^{2}$.Using $\pi \approx \frac{22}{7}$,we get $\vec{M}_{net} = -1.12 	imes \frac{22}{7} \hat{k} = -0.16 	imes 22 \hat{k} = -3.52 \hat{k} \, A \cdot m^{2}$.This is approximately $-\frac{7}{2} \hat{k} \, A \cdot m^{2}$.

Two concentric circular loops of radii $r_{1} = 30 \, cm$ and $r_{2} = 50 \, cm$ are placed in the $X-Y$ plane as shown in the figure. $A$ current $I = 7 \, A$ flows through them in the directions shown. The net magnetic moment of this system of two circular loops is approximately:

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