Question

A proton traveling at 27.1° with respect to the direction of a magnetic field of strength 2.33 mt experiences a magnetic force of 6.54 × 10-17 n. calculate (a) the proton's speed and (b) its kinetic energy in electron-volts.

Answer 1

(a) The magnetic force experienced by a charged particle is:
$F=qvB \sin \theta$
where
q is the charge
v is the velocity of the particle
B is the magnitude of the magnetic field
$\theta$ is the angle between the directions of v and B

The proton in our problem has a charge of $q=1.6 \cdot 10^{-19} C$, and it travels through a magnetic field with strength 
$B=2.33 mT=2.33 \cdot 10^{-3} T$
The direction between its velocity and B is $\theta=27.1 ^{\circ}$, and the force exerted on the proton is $F=6.54 \cdot 10^{-17}N$. Re-arranging the previous equation and using these data, we can find the value of v:
$v= \frac{F}{qB \sin \theta} = \frac{6.54 \cdot 10^{-17}N}{(1.6 \cdot 10^{-19} C)(2.33 \cdot 10^{-3}T)(\sin 27.1^{\circ})}=3.85 \cdot 10^5 m/s$

(b) Using the mass of the proton, $m=1.67 \cdot 10^{-27}kg$, we find its kinetic energy:
$K= \frac{1}{2} mv^2= \frac{1}{2}(1.67 \cdot 10^{-27} kg)(3.85 \cdot 10^5 m/s)^2=1.24 \cdot 10^{-16} J$

And keeping in mind that $1 eV = 1.6 \cdot 10^{-19}J$, we can convert this value into electronvolts:
$K= \frac{1.24 \cdot 10^{-16} J}{1.6 \cdot 10^{-19} J/eV}=775 eV$