A uniform,constant magnetic field vec B is di | vedclass

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Question

(A) The magnetic moment $\vec m$ of the square loop is given by $\vec m = I_0 A \hat n,$ where $A = L^2$ is the area and $\hat n$ is the unit vector n

Vedclass · Accepted Answer

$\vec 	au = \frac{B I_0 L^2}{\sqrt 2} (- \hat i + \hat j)$

Vedclass · Answer

(A) The magnetic moment $\vec m$ of the square loop is given by $\vec m = I_0 A \hat n,$ where $A = L^2$ is the area and $\hat n$ is the unit vector normal to the loop. Since the loop is in the $xy-$ plane,$\hat n = \hat k.$ Thus,$\vec m = I_0 L^2 \hat k.$The magnetic field $\vec B$ is in the $xy-$ plane at an angle of $45^o$ to the $x-$ axis,so $\vec B = B \cos 45^o \hat i + B \sin 45^o \hat j = \frac{B}{\sqrt 2} (\hat i + \hat j).$The torque $\vec 	au$ is given by $\vec 	au = \vec m 	imes \vec B.$$\vec 	au = (I_0 L^2 \hat k) 	imes \left( \frac{B}{\sqrt 2} (\hat i + \hat j) ight)$$\vec 	au = \frac{I_0 L^2 B}{\sqrt 2} (\hat k 	imes \hat i + \hat k 	imes \hat j)$Using the cross product rules $\hat k 	imes \hat i = \hat j$ and $\hat k 	imes \hat j = -\hat i,$$\vec 	au = \frac{I_0 L^2 B}{\sqrt 2} (\hat j - \hat i) = \frac{B I_0 L^2}{\sqrt 2} (- \hat i + \hat j).$

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