1.

When as proton has a velocity `v=(2hati+3hatj)xx10^6m/s`, it experience a force `F=-(1.28xx10^-13hatk)` When its velocity is along the z-axis, it experience a force along the `x`-axis. What is the magnetic field?

Answer» Substituting proper values in `F_m=q(vxxB)`
We have `-(1.28xx10^-13hatk)=(1.6xx10^-19)[(2hati+3hatj)xx(-B_0hatj)]xx10^6`
`:. 1.28=1.6xx2xxB_0`
or `B_0=1.28/3.2=0.4`
Therefore, the magnetic field is `B=(-0.4hatj)T`


Discussion

No Comment Found

Related InterviewSolutions