A magnet of magnetic moment 50 hat{i} text{ A | vedclass

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(B) The torque $\vec{	au}$ acting on a magnetic dipole in a magnetic field is given by the cross product: $\vec{	au} = \vec{\mu} 	imes \vec{B}$.Giv

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$150 \hat{k} 	ext{ Nm}$

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(B) The torque $\vec{	au}$ acting on a magnetic dipole in a magnetic field is given by the cross product: $\vec{	au} = \vec{\mu} 	imes \vec{B}$.Given:$\vec{\mu} = 50 \hat{i} 	ext{ Am}^2$$\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) 	ext{ T}$Substituting the values:$\vec{	au} = (50 \hat{i}) 	imes (0.5 \hat{i} + 3.0 \hat{j})$Using the distributive property of the cross product:$\vec{	au} = (50 	imes 0.5) (\hat{i} 	imes \hat{i}) + (50 	imes 3.0) (\hat{i} 	imes \hat{j})$Since $\hat{i} 	imes \hat{i} = 0$ and $\hat{i} 	imes \hat{j} = \hat{k}$:$\vec{	au} = 0 + 150 \hat{k} 	ext{ Nm}$$\vec{	au} = 150 \hat{k} 	ext{ Nm}$.

$A$ magnet of magnetic moment $50 \hat{i} \text{ Am}^2$ is placed along the $x$-axis in a magnetic field $\vec{B} = (0.5 \hat{i} + 3.0 \hat{j}) \text{ T}$. The torque acting on the magnet is:

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