A straight horizontal wire of length 0.50 m carries a current of 4.0 A from east to west. The wire lies in a uniform horizontal magnetic field of magnitude 0.20 T directed from north to south. What is the magnitude and direction of the magnetic force on the wire?
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Correct answer
A. 0.40 N upward
Principle or equation
The magnetic force on a current-carrying wire is given by F = B I L sin θ, where θ is the angle between the current direction and the magnetic field. The direction is given by the right-hand rule (or vector cross product).
Why this answer is correct
The current is east-west and the magnetic field is north-south, so the angle between them is 90° and sin θ = 1. Thus F = B I L = 0.20 T × 4.0 A × 0.50 m = 0.40 N. Using the right-hand rule: point fingers east (current), bend them toward south (field), the thumb points upward. So the force is 0.40 N upward.
Example
If a wire carrying 2.0 A along the x-axis is in a 0.50 T field along the y-axis, the force per unit length is F/L = B I = 1.0 N/m, directed along z (out of page).
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