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: $\vec{	au} = \vec{M} 	imes \vec{B}$.Given

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$150\,\hat{k}\,	ext{N}\,	ext{m}$

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(B) The torque $\vec{	au}$ acting on a magnetic dipole in a magnetic field is given by the cross product: $\vec{	au} = \vec{M} 	imes \vec{B}$.Given,$\vec{M} = 50\,\hat{i}\,	ext{A}\,	ext{m}^2$ and $\vec{B} = (0.5\,\hat{i} + 3.0\hat{j})\,	ext{T}$.Substituting these values into the formula:$\vec{	au} = (50\,\hat{i}) 	imes (0.5\,\hat{i} + 3.0\hat{j})$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 $\hat{i} 	imes \hat{j} = \hat{k}$:$\vec{	au} = 0 + 150\,\hat{k} = 150\,\hat{k}\,	ext{N}\,	ext{m}$.

$A$ magnet of magnetic moment $50\,\hat{i}\,\text{A}\,\text{m}^2$ 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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