A proton travels through uniform magnetic and electric fields. The magnetic field is in the negative x direction and has a magnitude of 2.35 mT. At one instant the velocity of the proton is in the positive y direction and has a magnitude of 2540 m/s. At that instant, what is the magnitude of the net force acting on the proton if the electric field is (a) in the positive z direction and has a magnitude of 4.19 V/m, (b) in the negative z direction and has a magnitude of 4.19 V/m, and (c) in the positive x direction and has a magnitude of 4.19 V/m?
Answers: 2
![F=(1.6*10^{-19}C)[4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=1.62*10^{-18}N](/tpl/images/0602/6368/021eb.png)
![F=(1.6*10^{-19}C)[-4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=2.84*10^{-19}N](/tpl/images/0602/6368/f00ea.png)
![F=(1.6*10^{-19}C)[4.19\frac{V}{m}\hat{i}+(2540\frac{m}{s})(2.35*10^{-3}T)\hat{k}]\\\\F=[6.704*10^{-19}\hat{i}+9.55*10^{-19}\hat{k}]N](/tpl/images/0602/6368/0b335.png)
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