A current-carrying circular loop of radius 2 | vedclass

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(B) The torque $\vec{	au}$ on a current loop is given by $\vec{	au} = \vec{m} 	imes \vec{B}$, where $\vec{m} = I \vec{A}$.Given $r = 2 	ext{ cm} =

Vedclass · Accepted Answer

$5024 	imes 10^{-7} \hat{k}$

Vedclass · Answer

(B) The torque $\vec{	au}$ on a current loop is given by $\vec{	au} = \vec{m} 	imes \vec{B}$, where $\vec{m} = I \vec{A}$.Given $r = 2 	ext{ cm} = 0.02 	ext{ m}$, the area $A = \pi r^2 = \pi (0.02)^2 = 4 	imes 10^{-4} \pi 	ext{ m}^2$.The magnetic moment vector is $\vec{m} = I A \hat{n} = (100\sqrt{2}) 	imes (4 	imes 10^{-4} \pi) 	imes \frac{\hat{i} + \hat{j}}{\sqrt{2}} = 4\pi 	imes 10^{-2} (\hat{i} + \hat{j}) 	ext{ A}\cdot	ext{m}^2$.The magnetic field is $\vec{B} = 4 	imes 10^{-3} (3\hat{i} + 2\hat{j}) = (12 	imes 10^{-3} \hat{i} + 8 	imes 10^{-3} \hat{j}) 	ext{ T}$.Calculating the cross product $\vec{	au} = \vec{m} 	imes \vec{B}$:$\vec{	au} = [4\pi 	imes 10^{-2} (\hat{i} + \hat{j})] 	imes [4 	imes 10^{-3} (3\hat{i} + 2\hat{j})]$$\vec{	au} = 16\pi 	imes 10^{-5} [(\hat{i} + \hat{j}) 	imes (3\hat{i} + 2\hat{j})]$$\vec{	au} = 16\pi 	imes 10^{-5} [\hat{i} 	imes 3\hat{i} + \hat{i} 	imes 2\hat{j} + \hat{j} 	imes 3\hat{i} + \hat{j} 	imes 2\hat{j}]$$\vec{	au} = 16\pi 	imes 10^{-5} [0 + 2\hat{k} - 3\hat{k} + 0] = 16\pi 	imes 10^{-5} (-\hat{k}) = -16 	imes 3.14 	imes 10^{-5} \hat{k} = -50.24 	imes 10^{-5} \hat{k} = -5024 	imes 10^{-7} \hat{k} 	ext{ N}\cdot	ext{m}$.The magnitude is $5024 	imes 10^{-7} 	ext{ N}\cdot	ext{m}$, which matches option $(B)$.

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