A uniform magnetic field of 2 times 10^{-3} | vedclass

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(B) The area vector $\vec{A}$ of the loop in the $Y-Z$ plane,with current flowing anticlockwise when viewed from the negative $X$-axis,is directed alo

Vedclass · Accepted Answer

$2 	imes 10^{-4} \ N-m$ along the negative $Z$-direction

Vedclass · Answer

(B) The area vector $\vec{A}$ of the loop in the $Y-Z$ plane,with current flowing anticlockwise when viewed from the negative $X$-axis,is directed along the negative $X$-axis: $\vec{A} = (0.2 \ m 	imes 0.1 \ m)(-\hat{i}) = 0.02(-\hat{i}) \ m^2$.The magnetic moment $\vec{M}$ is given by $\vec{M} = i\vec{A} = 5 \ A 	imes 0.02(-\hat{i}) \ m^2 = 0.1(-\hat{i}) \ A-m^2$.The magnetic field is $\vec{B} = 2 	imes 10^{-3} \hat{j} \ T$.The torque $\vec{	au}$ is given by $\vec{	au} = \vec{M} 	imes \vec{B}$.$\vec{	au} = [0.1(-\hat{i})] 	imes [2 	imes 10^{-3} \hat{j}] = 0.1 	imes 2 	imes 10^{-3} 	imes (-\hat{i} 	imes \hat{j}) = 2 	imes 10^{-4} 	imes (-\hat{k}) \ N-m$.Thus,the torque is $2 	imes 10^{-4} \ N-m$ along the negative $Z$-direction.

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