Question Detail
Question
A coil has 45 turns and an area of 1.8 × 10-2m2. If the torque on the coil is 2.5 × 10-2 Nm when placed in a magnetic field of 0.08T, calculate the current flowing in the coil.
Correct Answer
Option B is the correct answer.
Detailed Explanation
The torque (τ) on a current-carrying coil in a magnetic field is given by the formula:τ = N I A B sin(θ)Where:N = number of turns = 45I = current (what we need to find)A = area of the coil = 1.8 × 10-2 m2B = magnetic field strength = 0.08 Tθ = angle between the magnetic field and the normal to the coil's area. For maximum torque, θ = 90°, so sin(θ) = 1.Given τ = 2.5 × 10-2 Nm (maximum torque, assuming sin(θ)=1)I = τ / (N A B)I = (2.5 × 10-2 Nm) / (45 * 1.8 × 10-2 m2 * 0.08 T)I = 0.025 / (45 * 0.018 * 0.08)I = 0.025 / 0.0648I ≈ 0.3858 ARounding to two significant figures, I ≈ 0.39 A.
Hint
Use the formula for torque on a current loop in a magnetic field, assuming maximum torque occurs when the field is perpendicular to the area vector of the coil.