+3 votes
in JEE by kratos

A charge particle (Q = 10-4 C) is released from rest at z = 0 in magnetic field given as bar B = B0 cos(ωt - kz) Î + B1 cos((ωt - kz)j where B0 = 3 x 10-5 and B1 = 2 x 10-6. Then the rms value of force acting on particle is ?

(1) 3 x 10-2

(2) 0.6

(3) 0.9

(4) 0.1

1 Answer

+1 vote
by kratos
 
Best answer

Correct option: (2) 0.6

Explanation:

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