When a proton has a velocity
`v=(2hati+3hat)xx10^(6)ms^(-1)`, ti experiences a force `F=-(1.28xx10^(-13)hatk)N.` When its velocity is along the Z-axis then 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`.