A 0.300 m horizontal wire segment of mass 18.0 g lies perpendicular to a uniform horizontal magnetic field of 0.250 T. The wire is supported so that magnetic force may act vertically. Determine the current magnitude and direction needed for the magnetic force to balance its weight. If the current is 20% larger, find the initial vertical acceleration. Discuss why the complete circuit geometry matters and how a balance could measure B.

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Required current = 2.35 A, directed so I × B is upward. At 20% larger current the initial acceleration is 1.96 m s⁻² upward. For a straight segment perpendicular to B, magnetic force magnitude is F_B = BIL. Balance requires BIL = mg. I = mg/(BL) = (0.0180)(9.81)/[(0.250)(0.300)] = 2.35 A. Use the vector rule I × B to choose whichever wire direction makes force upward. At 1.20I_balance, magnetic force is 1.20mg. Net upward force is 0.20mg, so initial acceleration is 0.20g = 1.96 m s⁻². Other circuit segments can experience magnetic forces too. Their directions, lengths within the field and mechanical supports determine whether they contribute to the measured vertical force or cancel. Place the field segment on a balance, reverse current to remove constant offsets, and use half the difference in readings. A graph of force against IL has gradient B when the segment is perpendicular; control temperature because wire resistance and current can drift.

Balance a current-carrying wire

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