Practice question
Question
A straight wire of length \( 1.2 \, \text{m} \) carries a current of \( 6 \, \text{A} \) perpendicular
to a uniform magnetic field of \( 0.25 \, \text{T} \). What is the force acting on the wire?
Explanation
**Field outside long solenoid** considered negligible because magnetic lines are concentrated inside and return path spreads over large area outside, making B_out ≈ 0. This justifies using solenoid for uniform field experiments, with n = 950 m⁻¹, I = 1.4 A giving B = 4π×10⁻⁷×950×1.4 = 1.67×10⁻³ T. Force F = I l B sin θ , where θ = 90° , so sin θ = 1 . F = 6 × 1.2 × 0.25 = 1.8 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ =
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