When a proton has a velocity `vec v = (2hat i + 3 hat j) xx 10^6 ms^-1`, it experiences a force `vec F = -(1.28 xx 10^(-13) hat k) N.` When its velocity is along the z-axis, it experiences a force along the x-axis. What is the magnetic field?

Explanation

(a) In first case, force is along `-hatk`, so `vecB` should be in xy plane
or z component of B should be zero. In second case, force is along x axis, so B can not be along x-axis. Finally we find that B is along negative y direction.
Let `vecB=-B_0hat j` and applying `vecF=qvec v xx vec B`
`-(1.28xx10^-13hatk)=(1.6xx10^-19)xx[(2hati+3hatj)xx(-B_0hatj)]xx10^6`
`1.28=1.6xx2xxB_0` or `B_0=0.4T`
`vecB=-(0.4hatj)T`.