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Question

A particle with a charge of 0.02 C is moving at right angles to a uniform magnetic field with a strength of 0.3 T. The velocity of the charge is 740 m/s. What is the magnitude of the magnetic force exerted on the particle?
_____ N

Answer 1

F= q B x v
F- force
Q – charge
B magnetic field intensity
v – velocity of the charge
x- designates a vector product


Or since the magnetic field is normal to the velocity of the charge we have
F=q B v
F=0.02 x 0.3 x 740
F=4.44N

Edit 1
As suggested by anilbakshi in this problem the charge sign convention is important as it signifies the direction of the resulting force and therefore using –q and of +q is required.