First slide
Force on a charged particle moving in a magnetic field
Question

A proton (mass =1.67×1027kg and charge =1.6×1019C is projected in a uniform magnetic field of 2 Wb/m2 with a velocity 3.4×107m/s in a direction perpendicular to the field. The acceleration of the proton should be

Moderate
Solution

F = ma = qvB

a =qvBm=1.6×1019×2×3.4×1071.67×1027

=6.5×1015 m/s2

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