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#perpendicular field

2 public questions tagged with this topic.

Magnetic Force on Moving Charge - Lorentz Force and Motion

A wire of length \( 2 \, \text{m} \) carries a current of \( 5 \, \text{A} \) and is placed perpendicular to a magnetic

**Motion of charged particle perpendicular to uniform B** is circular because force F ⊥ v, no work done, speed constant, radius r = m v/(q B), period T = 2π m/(q B) independent of v. If v has component parallel to B, helical path results, pitch = v_parallel·T. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 5 × 2 × 0.4 = 4 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =

Ref: NCERT > Physics Book > Moving Charge and Magnetism > Magnetic Force on Moving Charge - Lorentz Force and Motion

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Torque on Magnetic Dipole and Potential Energy

A dipole with \( m = 0.25 \, \text{A m}^2 \) in a field \( B = 0.4 \, \text{T} \) at \( 90^\circ \) has potential energy

**Magnetic dipole in uniform field** experiences torque τ = m B sinθ and potential energy U = -m·B = -m B cosθ, minimum -mB when aligned (θ=0°), maximum +mB at anti-alignment (θ=180°). Work done rotating from θ₁ to θ₂ equals ΔU = mB(cosθ₁ - cosθ₂). U_m = -m B cosθ . Given: m = 0.25 A m² , B = 0.4 T , θ = 90° , cos 90° = 0 . U_m = -0.25 × 0.4 × 0 = 0 J . Substituting values gives 0 J, which matches expected magnitude for this magnetic configuration, confirming dipole field dependence on m/r³ and torque relation τ = m B sinθ.

Ref: NCERT > Physics Book > Magnetism and Matter > Torque on Magnetic Dipole and Potential Energy

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