Question

Through what potential difference would the deuteron have to be accelerated to acquire this speed?

Answer 1

Under the action of magnetic field,

$\frac{m{v}^2}{r}=qvB$

$⟹mv=qBr$

Hence kinetic energy=$\frac{1}{2}m{v}^2=\frac{q^2{B}^2{r}^2}{2m}$

This kinetic energy is achieved by application of outside potential difference.

Hence $qV=\frac{q^2{B}^2{r}^2}{2m}$

$⟹V=7.26kV$