Answer:
Velocity of the proton, [tex]v_{p} = 250 m/s[/tex]
Given:
Magnetic field, B = 2 T
Electric field, E = 500 V/m
Explanation:
For the proton to travel in a straight line without any change in direction in both the electric and magnetic field, the forces on the proton in both the electric and magnetic fields must be equal and is given by:
[tex]F_{E} = F_{M}[/tex]
where
[tex]F_{E} = Force on proton in electric field[/tex]
[tex]F_{M} = Force on proton in magnetic field[/tex]
Now,
[tex]q_{p}E = q_{p}(\vec{v_{p}}\times \vec{B})[/tex]
where
[tex]q_{p} = charge of proton[/tex]
[tex]v_{p} = velocity of proton[/tex]
E = electric field
[tex]q_{p}E = q_{p}v_{p}Bsin\theta = q_{p}v_{p}Bsin90^{\circ}[/tex]
(Since, [tex]sin90^{\circ} = 1[/tex])
[tex]v_{p} = \frac{E}{B}[/tex]
[tex]v_{p} = \frac{500}{2} = 250 m/s[/tex]
