Question

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?

Answer 1

Answer:

a) 1.62*10^{-18}N

b) 2.84*10^{-19}N

c) 1.16*10^{-18}N

Explanation:

The net force is given by the expression:

$\vec{F}=q(\vec{E}+\vec{v}\ X\ \vec{B})$

By the right hand rule, we know that the direction of the magnetic force is

j X -i = k

$\vec{v}X\vec{B}=vB\hat{k}$

Hence, the direction of the magnetic force is the +z direction.

(a) E=4.19V/m k

$F=(1.6*10^{-19}C)[4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=1.62*10^{-18}N$

where we have used q=9.1*10^{-31}kg.

(b) E=-4.19V/m k

$F=(1.6*10^{-19}C)[-4.19\frac{V}{m}+(2540\frac{m}{s})(2.35*10^{-3}T)]=2.84*10^{-19}N$

(c) E=4.19V/m i

In this case the net force will have two components:

$F=(1.6*10^{-19}C)[4.19\frac{V}{m}\hat{i}+(2540\frac{m}{s})(2.35*10^{-3}T)\hat{k}]\\\\F=[6.704*10^{-19}\hat{i}+9.55*10^{-19}\hat{k}]N$

and its magnitude will be:

$F=\sqrt{(6.704*10^{-19})^2+(9.55*10^{-19})^2}=1.16*10^{-18}N$

hope this helps!!