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Torque on a current loop
Appears in
Concepts tested here
- Angle between area vector and field
- Equilibrium of a magnetic dipole
- Magnetic moment of a solenoid
All Questions
2015 AIPMT-II 1 question
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A rectangular coil of length 0.12 m and width 0.1 m having 50 turns of wire is suspended vertically in a uniform magnetic field of strength 0.2 Weber/$m^2$. The coil carries a current of 2 A. If the plane of the coil is inclined at an angle of $30^\circ$ with the direction of the field, the torque required to keep the coil in stable equilibrium will be:The plane makes $30^\circ$ with $\vec B$, so the area vector (normal) makes $\theta = 60^\circ$ with $\vec B$.
$\vec\tau = \vec M\times\vec B$, $|\tau| = NIAB\sin\theta$
$|\tau| = 50\times 2\times(0.12\times0.1)\times 0.2\times\sin 60^\circ \approx 0.20$ Nm
2013 NEET 1 question
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A current loop in a magnetic field:The potential energy of the loop is $U = -MB\cos\theta$.
It is in equilibrium when $\vec M$ is parallel to $\vec B$ ($\theta = 0^\circ$, stable) and when it is antiparallel ($\theta = 180^\circ$, unstable).
2010 AIPMT-MAINS 1 question
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A closely wound solenoid of 2000 turns and area of cross-section $1.5\times10^{-4}\ m^2$ carries a current of 2.0 A. It is suspended through its centre and perpendicular to its length, allowing it to turn in a horizontal plane in a uniform magnetic field $5\times10^{-2}$ tesla making an angle of $30^\circ$ with the axis of the solenoid. The torque on the solenoid will beMagnetic moment of the solenoid: $M = NIA = 2000\times2.0\times1.5\times10^{-4} = 0.6\ A\,m^2$
The magnetic moment is along the axis of the solenoid, which makes $30^\circ$ with the field.
Torque: $\tau = MB\sin\theta$
$\tau = 0.6\times5\times10^{-2}\times\sin30^\circ = 0.6\times5\times10^{-2}\times\frac{1}{2}$
$\tau = 1.5\times10^{-2}$ N m