Question

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?

Answer 1

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

v=veloicty vector

B=magnetic field vector

given that,

q=1.6*10^(-19) C
v=4*10^4 j m/s

B=-0.2 i

then force F=1.6*10^(-19)*cross product of (4*10^4 j and -0.2 i)

=1.6*10^(-19)*(8*10^3 k)

=1.28*10^(-15) k

hence magnitude of force is 1.28*10^(-15) N

and it is directed towards +ve z axis.