91Ó°ÊÓ

$$\begin{gathered} âˆ\dots =23.0^0 \\ B=2.60mT=2.60×{10}^{-3}T \\ {F}_B=6.50×{10}^{-17N} \end{gathered}$$

Find the velocity of the proton using the formula for magnetic field in terms of charge and velocity of proton. Then using the formula for K.E, find the kinetic energy of the proton.

Formulae are as follow:

$$\begin{gathered} F_B=evB\sin âˆ\dots \\ K.E.=\frac{1}{2}m{v}^2 \end{gathered}$$

Where, F_B is magnetic force, v is velocity, m is mass, K.E. is kinetic energy, B is magnetic field, e is charge of particle.

The magnetic force experienced by the proton is,

$F_B=evB\sin âˆ\dots$

Hence, velocity of proton is,

$$\begin{gathered} v=\frac{F_B}{eB\sin âˆ\dots } \\ v=\frac{6.50×{10}^{-17}}{\left(1.6×{10}^{-19}\right)\left(2.60×{10}^{-3}\right)\sin \left(23.0^0\right)} \\ v=399891~4.00×{10}^5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right. \end{gathered}$$

Hence, the velocity of proton is $4.00×{10}^5\raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$.

Kinetic energy of the proton is,

$$\begin{gathered} K.E=\frac{1}{2}m{v}^2 \\ K.E=\frac{1}{2}\left(1.67×{10}^{-27}\right){\left(399891\right)}^2 \\ K.E=1.3352×{10}^{-16}J \\ K.E=\frac{1.3352×{10}^{-16}}{1.6×{10}^{-19}} \\ K.E=834.5~835eV \end{gathered}$$

Hence, the kinetic energy of the proton is 835 eV .

Therefore, by using the formula for magnetic field in terms of charge and velocity of proton and the formula for K.E, the velocity and kinetic energy of proton can be determined.