Question

An electron and a proton are injected into a uniform magnetic field perpendicular to it in the same direction. If electron and proton have same kinetic energy then the radius of curvature is

• A. more for proton
• B. more for electron
• C. same for both
• D. none of thesr

Answer 1

Answer: (A)

more for proton

let, Kinetic energies of electron and proton be K.
magnetic force on particle provides the centripetal force to support the circular motion.
If r is radius of curvature, $F_b=F_c$
i.e. qvB = m $\frac{v^2}{r}$.
Solving the above equation we get r = $\frac{p}{qB}$ where p = momentum = mv
If, q and B are constant we can say that $r\alpha p$
But, K.E. = $\frac{m{v}^2}{2}$.
Derive P in terms of K.E. which turns out to be
p = ${\left(2m\right(K.E.\left)\right)}^{\frac{1}{2}}$
Given that, K.E. is same for either particles. So, $p\alpha m^{\frac{1}{2}}$
therefore, $r\alpha p\alpha m^{\frac{1}{2}}$ and radius is more for higher mass particle which is a proton in this case.