The force on a current-carrying wire in a magnetic field is the strongest when

Question

BlackTom AI · Accepted Answer

The magnetic force on a straight current-carrying wire has magnitude F = I L B sinθ, so the angle θ between the current direction (wire) and the magnetic field determines the force.

1) Relevant knowledge points and concepts
- Lorentz force for a current element: dF = I (dL × B). For a straight wire of length L, magnitude becomes F = I L B sinθ.
- I is current, L is length, B is magnetic field magnitude, θ is the angle between the wire (current direction) and the magnetic field.
- The sine function ranges from 0 to 1; force is proportional to sinθ.

2) Step-by-step reasoning
- Use F = I L B sinθ to compare forces at different angles.
- For each option compute sinθ and hence the relative force (factor times I L B).

3) Calculations (intermediate steps)
- θ = 0° (current parallel to field): sin0° = 0 → F = 0 · I L B = 0.
- θ = 30°: sin30° = 1/2 = 0.5 → F = 0.5 · I L B.
- θ = 60°: sin60° = √3/2 ≈ 0.866 → F ≈ 0.866 · I L B.
- θ = 90° (perpendicular): sin90° = 1 → F = 1 · I L B = I L B (maximum).

4) Why the correct answer is correct
- Since sinθ reaches its maximum value 1 at θ = 90°, the magnitude F = I L B sinθ is largest when the current is perpendicular to the magnetic field lines.

5) Why other options are incorrect
- Parallel (0°): gives sin0 = 0, so no magnetic force — incorrect.
- 30°: gives only 50% of the maximum force — less than perpendicular.
- 60°: gives ≈86.6% of maximum — still less than perpendicular.
- Only the perpendicular case yields the full I L B force.

6) Conclusion
- Therefore the force is strongest when the current is perpendicular to the field lines. This matches the provided correct answer: "the current is perpendicular to the field lines."

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