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?
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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`
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