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In the figure to the right, a proton moves into a region of uniform magnetic field of magnitude 0.25 T, goes through half a cPlease answer both questions

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Answer #1

2 T = NIAB Sino where N 300 I=5A B = 0.85T 0 = 30° A = 0.02X0-035 So T = 300x5x(0.02X0:035) X (300Xšinu Hence the torque is P

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