A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50 T experiences a force of what magnitude?

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To determine the magnetic force on a current-carrying wire in a magnetic field, use F = I L B sin(theta), where theta is the angle between the wire (direction of current) and the magnetic field.

Option 1: 6.7 N. This is far larger than the maximum possible force for the given values. The maximum magnitude occurs when sin(theta) = 1 (wire perpendicular to field), giving F_max = I L B = (0.60 A)(2.0 m)(0.50 T) = 0.60 N. Therefore, 6.7 N is not achievable with the provided data.

Option 2: 0.30 N. This would correspond to sin(theta) = F/(I L B) = 0.30 N / 0.60 N = 0.5, implying theta ≈ 30 degrees. However, the problem states the wire is oriented parallel to the magnetic field, which means theta = 0 degrees and sin(theta) = 0, so the force should be zero.

Option 3: 0.15 N. This would imply sin(theta) = 0.25, corresponding to theta ≈ 14.5 degrees. Again, the given orientation is parallel, so such an intermediate angle is not consistent with the scenario described.

Option 4: 0 N. With the wire parallel to the magnetic field, theta = 0° and sin(0) = 0, giving F = I L B sin(0) = 0. Hence the force magnitude is zero.

APPHYSC142-031 with labs Exam Review - Final

A 2.0 m wire segment carrying a current of 0.60 A oriented parallel to a uniform magnetic field of 0.50 T experiences a force of what magnitude?

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