A proton moving with a velocity of 4.0×10^4 m/s along the +y-axis enters a magnetic field of 0.20 T directed towards the -x-axis. What is the magnitude force acting on the proton?
i,j and k are unit vectors along +ve x axis, +ve y axis, +ve z axis respectively.
as we know, magnetic force on a charged particle is given by
F=q*(cross product of v and B)
where q=charge on the particle
B=magnetic field vector
v=4*10^4 j m/s
then force F=1.6*10^(-19)*cross product of (4*10^4 j and -0.2 i)
hence magnitude of force is 1.28*10^(-15) N
and it is directed towards +ve z axis.
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