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  • A proton moving with a speed u in the x-y plane along the positive x axis enters at y=0 a region of uniform magnetic field B directed into the x-y plane as shown in Fi

A proton moving with a speed u in the x-y plane along the positive x axis enters at y=0 a region of uniform magnetic field B directed into the x-y plane as shown in Fig. After sometime, the proton leaves the region with a speed v at co-ordinate y. Then

Option: 1

v = u, y > 0


Option: 2

v = u, y < 0


Option: 3

v > u, y > 0


Option: 4

v > u, y < 0


Answers (1)

best_answer

F = q (u x B)} . Since force F is perpendicular to u, it does no work on the particle. Hence the speed of the proton remains unchanged, i.e. v = u [see Fig.]
From Fleming's L.H. rule, the force is directed upwards. Hence the proton, after completing a semicircle in the region of magnetic field emerges at positive y-coordinate.
So the correct choice is (a).

Posted by

sudhir.kumar

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