(D) The torque on a current-carrying coil in a uniform magnetic field is given by $\vec{\tau} = \vec{m} \times \vec{B}$,where $\vec{m}$ is the magneti

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perpendicular to the magnetic field

(D) The torque on a current-carrying coil in a uniform magnetic field is given by $\vec{\tau} = \vec{m} \times \vec{B}$,where $\vec{m}$ is the magnetic moment and $\vec{B}$ is the magnetic field. The coil experiences zero torque when the magnetic moment $\vec{m}$ is parallel to the magnetic field $\vec{B}$. The magnetic moment vector $\vec{m}$ is always perpendicular to the plane of the coil. For $\vec{m}$ to be parallel to $\vec{B}$,the plane of the coil must be perpendicular to the magnetic field $\vec{B}$.

Class 12 Physics Moving Charges and Magnetism Torque , Potential Energy and Work Done in Mangetic Field

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$A$ current-carrying coil is subjected to a uniform magnetic field. The coil will orient itself so that its plane becomes

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A
inclined at $45^{\circ}$ to the magnetic field
B
inclined at any arbitrary angle to the magnetic field
C
parallel to the magnetic field
D
perpendicular to the magnetic field

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Moving Charges and Magnetism

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Torque , Potential Energy and Work Done in Mangetic Field

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$A$ current-carrying coil is subjected to a uniform magnetic field. The coil will orient itself so that its plane becomes

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