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A straight conducting rod PQ of length 5 m is oriented along the y-axis and moves with velocity v = 2 m/s in the +x direction (no rotation) in a uniform magnetic field B = (3 j-hat + 4 k-hat) Tesla. What is the EMF induced in the rod?

1. 0 V
2. 10 V
3. 20 V
4. 40 V

Correct answer: 20 V

Solution

v = 2 x-hat m/s, B = 3 y-hat + 4 z-hat T. v cross B = (2 x-hat) cross (3 y-hat + 4 z-hat) = 2*3*(x cross y) + 2*4*(x cross z) = 6 z-hat + 8(-y-hat) = -8 y-hat + 6 z-hat. The rod is along the y-axis, so dl = dy y-hat. Component of (v cross B) along y = -8 V/m. EMF = integral from 0 to 5 of (-8) dy = -8*5 = -40 V? Magnitude = 40 V. But wait: x-hat cross z-hat = -(z cross x) = -y-hat. So v cross B = 6 z-hat + 8*(-y-hat) = -8 y-hat + 6 z-hat. Component along rod (y-axis) = -8. EMF = 8*5 = 40 V. Hmm, but that gives option D. Let me recheck: If rod is along y-axis, l-hat = y-hat, length = 5 m. EMF = (v cross B). l = (-8 y-hat + 6 z-hat). (5 y-hat) = -8*5 = -40 V. |EMF| = 40 V. But option D is 40 V. However if the rod is oriented differently... if the rod is along the x-axis or z-axis, the answer changes. The original says 'oriented as shown' (missing figure). If rod is along z-axis: (v cross B). l-hat = (-8 y-hat + 6 z-hat). z-hat = 6; EMF = 6*5 = 30 V (not in options). If rod along y-axis: 40 V. If rod along x-axis: (v cross B).x-hat = 0; EMF = 0 (option A). Without figure, the standard JEE answer for this problem with rod along y-axis is 40 V. But the answer given in many solutions is 20 V, suggesting rod length contribution or rod oriented at 45 degrees. Given options include 20 V, let me try: if B = 3 j + 4 k and rod is along some direction giving 20 V... if rod is at angle, or if only B = 4 k-hat matters: v cross (4 k-hat) = 2 x cross 4 z = 8(x cross z) = -8 y-hat; EMF along y-rod = 8*5=40. If only 3 j matters: v cross (3 j) = 6(x cross y) = 6 z; component along y = 0. So with rod along y-axis the answer is 40 V. Given the most natural reading and options, answer is 40 V.

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