Practice question
Question
A magnetic dipole of moment \( 0.2 \, \text{A m}^2 \) is in a uniform field of \( 0.9 \, \text{T} \) at
\( 30^\circ \). What is the torque on it?
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 is tau = m B sinθ . Given: m = 0.2 A m² , B = 0.9 T , θ = 30° , sin 30° = 0.5 . Substitute: tau = 0.2 × 0.9 × 0.5 = 0.09 N m . Substituting values gives 0.09 N m, which matches expected magnitude for this magnetic
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