Practice question
Question
A wire of length \( 1 \, \text{m} \) with \( 10 \, \text{A} \) is at \( 45^\circ \) to a field of \(
0.2 \, \text{T} \). What is the force on it?
Explanation
**Force on current-carrying wire** in magnetic field is F = I l × B, magnitude F = I l B sinθ, I current (A), l length (m), B field (T), θ angle between current direction and B. Direction perpendicular to plane containing wire and B, given by Fleming's left-hand rule. F = I l B sin θ . F = 10 × 1 × 0.2 × sin 45° = 2 × (1/√(2)) = 1.414 ≈ 1.41 N . Using F = q v B sinθ, F = I l B sinθ, B = μ₀ I/(2π r), B = μ₀ N I/(2R) and τ = N
Discussion
Comments
Please log in to join the discussion.
Login to commentNo comments yet. Be the first to start the discussion.