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Question

A magnetic needle with magnetic moment \( 0.1 \, \text{A m}^2 \) is placed in a uniform magnetic field
of \( 0.2 \, \text{T} \) at an angle of \( 60^\circ \) with the field. What is the torque acting on it?

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Explanation

**Earth's magnetism** approximated as dipole inclined to rotation axis, magnetic declination is angle between geographic north and magnetic north, inclination or dip angle is angle between total field and horizontal. Horizontal component B_H = B cosδ, vertical B_V = B sinδ, δ dip angle, B_H ≈ 3-4×10⁻⁵ T in India. Torque on a magnetic dipole is tau = m B sinθ . Given: m = 0.1 A m² , B = 0.2 T , θ = 60° , sin 60° = (√(3)/2) ≈ 0.866 . Substitute: tau = 0.1 × 0.2 × 0.866 = 0.01732 N m ≈ 0.017 N m . Substituting values gives

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