Answer :
Answer:
[tex]r=1.14m[/tex]
Explanation:
[tex]\theta[/tex] is the angle between the velocity and the magnetic field. So, the magnetic force on the proton is:
[tex]F_m=qvBsen\theta\\F_m=qvBsen(90^\circ)\\F_m=qvB[/tex]
A charged particle describes a semicircle in a uniform magnetic field. Therefore, applying Newton's second law to uniform circular motion:
[tex]F_m=F_c\\qvB=F_c(1)[/tex]
[tex]F_c[/tex] is the centripetal force and is defined as:
[tex]F_c=m\frac{v^2}{r}[/tex]
Here [tex]v[/tex] is the proton's speed and [tex]r[/tex] is the radius of the circular motion. Replacing this in (1) and solving for r:
[tex]qvB=\frac{mv^2}{r}\\r=\frac{mv^2}{qvB}\\r=\frac{mv}{qB}[/tex]
Recall that 1 J is equal to [tex]6.242*10^{12}MeV[/tex], so:
[tex]4.9MeV*\frac{1J}{6.242*10^{12}MeV}=7.85*10^{-13}J[/tex]
We can calculate [tex]v[/tex] from the kinetic energy of the proton:
[tex]K=\frac{mv^2}{2}\\\\v=\sqrt{\frac{2K}{m}}\\v=\sqrt{\frac{2(7.85*10^{-13}J)}{1.67*10^{-27}kg}}\\v=3.06*10^{7}\frac{m}{s}[/tex]
Finally, we calculate the radius of the proton path:
[tex]r=\frac{mv}{qB}\\r=\frac{1.67*10^{-27}kg(3.06*10^{7}\frac{m}{s})}{1.6*10^{-19}C(0.28T)}\\r=1.14m[/tex]