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A proton moving in the XY plane with a velocity of 6×10^7m/s makes an angle of 30° with a uniform magnetic field of induction B=1.5T acting the positive direction of Y-axis. Find the force acting upon it.
Use formula F = qvB sinx. Here x= 30°.Substitute the values given in the question.U will gee the answer.Use q = 1.6×10^-19 C. Hoping that it will help u.RegardsRohit.
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