A proton is moving with a speed of $v = c/200$ ($c$ is the speed of light) in a uniform electric fie

Question

A proton is moving with a speed of $v = c/200$ ($c$ is the speed of light) in a uniform electric field $\vec{E}$ of magnitude $4.5 	imes 10^3 \mathrm{~V/m}$. To ensure the proton travels undeflected, a uniform magnetic field $\vec{B}$ must be applied perpendicular to both its velocity and the electric field. The magnitude of this required magnetic field is ($c = 3 	imes 10^8 \mathrm{~m/s}$)

Professor PC Thomas Classes · Answer

(c) :  For a charged particle to move undeflected through mutually perpendicular electric and magnetic fields, the electric force must perfectly balance the magnetic force. Thus, $qE = qvB$, which simplifies to $E = vB$. The speed of the proton is $v = \frac{3 	imes 10^8}{200} = 1.5 	imes 10^6 \mathrm{~m/s}$. The required magnetic field is $B = \frac{E}{v} = \frac{4.5 	imes 10^3}{1.5 	imes 10^6} = 3 	imes 10^{-3} \mathrm{~T}$.