A 4.0 cm long straight wire carrying a current of 8 A is placed perpendicular to a uniform magnetic field of strength 0.15 T . The magnetic force on the wire is ________mN .

Solution

Physics sir · 3 steps

Wire at right angles to the field, so the force is just BIL. Put length in metres, then convert to mN.

1. Force on a wire. A straight wire of length L carrying current I in a field B feels F = BIL sin θ. Here the wire is perpendicular to the field, so θ = 90° and sin θ = 1.

   F = BIL = 0.15 × 8 × 0.04

2. Put in numbers. Convert the length first: 4.0 cm = 0.04 m. Then 0.15 × 8 = 1.2, and 1.2 × 0.04 = 0.048 N.
3. So. F = 0.048 N = 48 mN, so the answer is 48.

The common slip: 0.048. Stopping at newtons gives 0.048, not 48. Multiply by 1000 to convert N to mN.

The idea to keep. For a straight wire in a uniform field, F = BIL sin θ, and it is largest when the wire is perpendicular to B. Always put L in metres, then convert the force to the unit asked.

Answer from the NTA final answer key. Solution by PrepWiser.