A straight conductor of length 0.40 m carrying a current of 5.0 A is positioned in a uniform magnetic field of 0.60 T perpendicular to the field lines. What magnetic force acts on the wire?
Answer and explanation
Correct answer: A. 1.20 N
- Step 1: Apply the magnetic force formula on a current-carrying wire: F = I L B * sin(theta).
- Step 2: Since the wire is perpendicular to the field, theta = 90° and sin(90°) = 1.
- Step 3: Compute force: F = (5.0 A) (0.40 m) (0.60 T) 1 = 2.0 0.60 = 1.20 N.
Answer
1.20 N.
Why not the others: '2.40 N' doubles the length or current; '0.60 N' halves the current; '12.0 N' misplaces the decimal place.
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