How Torque on a Bar Magnet Placed in Uniform Magnetic Field

How Torque on a Bar Magnet Placed in Uniform Magnetic Field

Consider a bar magnet NS of length 2l and pole strength m placed in a uniform magnetic field of induction B at an angle θ with the direction of the field (Figure). A uniform magnetic field can be prepared by making a comparatively long cylindrical coil.

Due to the magnetic field B, a force mB acts on the North Pole along the direction of the field and a force mB acts on the South Pole along the direction opposite to the magnetic field. The space or region around a magnet or a wire carrying current whereby the magnetic needle shows deflection, that space or region is called the magnetic field of that magnet or wire carrying current.

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These two forces are equal and opposite, hence constitute a couple.

The torque τ due to the couple is

τ = one of the forces x perpendicular distance between them

τ = F x NA

= mB x NA … …. (1)

= mB × 2l sin θ

so, τ = MB sin θ

Vectorially, τ = M x B

The direction of τ is perpendicular to the plane containing M and B.

If, B = 1 and θ = 90 θ

Then from equation (2), τ = M

Hence, the moment of the magnet M is equal to the torque necessary to keep the magnet at right angles to a magnetic field of unit magnetic induction.

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