A straight wire of length 0.40 m carries a current of 5.0 A and is placed in a uniform magnetic field of 0.30 T. The wire is oriented at an angle of 30° to the direction of the magnetic field. What is the magnitude of the magnetic force on the wire? (Use sin 30° = 0.5)
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Correct answer
A. 0.30 N
Principle or equation
The force on a current-carrying wire in a magnetic field is given by F = B I L sin θ, where θ is the angle between the wire and the magnetic field direction.
Why this answer is correct
Substitute the given values: F = 0.30 T × 5.0 A × 0.40 m × sin 30° = 0.30 × 5.0 × 0.40 × 0.5 = 0.30 N.
Example
If a 0.2 m wire carrying 3 A is perpendicular to a 0.5 T field, the force is 0.5 × 3 × 0.2 = 0.3 N.
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