A proton traveling at$\mathbf{\text{23.0°}}$with respect to the direction of a magnetic field of strength$\text{2.60}$mT experiences a magnetic force of${\mathbf{\text{6.50×10}}}^{\mathbf{\text{-17}}}$N.

(a) Calculate the proton’s speed

. (b)Find its kinetic energy in electron-volts.

$$\begin{gathered} f=23.0° \\ B={\text{2.60mT=2.60×10}}^{\text{-3}}\text{T} \\ {F}_B{\text{=6.50×10}}^{\text{-17}}\text{N} \end{gathered}$$

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:

$$\begin{gathered} F_B=evBsinf \\ K.E=\frac{1}{2}m{v}^2 \end{gathered}$$

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,

$F_B=evBsinf$

Hence, velocity of proton is,

$v=\frac{F_B}{eBsinf}$

$v=\frac{6.50\times {10}^{-17}}{(1.6\times {10}^{-19})(2.60\times {10}^{-3})\text{sin}(23.0°)}$

$v=399891{\text{~4.00×10}}^{\text{5}}\text{m/s}$

Hence, the velocity of proton is.${\text{4.00×10}}^{\text{5}}\text{m/s}$

Kinetic energy of the proton is,

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

Hence, thekinetic energy of the proton is.$\text{835eV}$

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.