Question

A beam of protons moves in a circle of radius 0.20 m. The protons move perpendicular to a 0.36-T magnetic field. (a) What is the speed of each proton? (b) Determine the magnitude of the centripetal force that acts on each proton.

Answer 1

Answer:

a) $v=6.898\times 10^{6}\ m.s^{-1}$ is the speed of each proton

b) $F_c=3.97\times 10^{-13}\ N$

Explanation:

Given:

radius of path of motion, $r=0.2\ m$

we know charge on protons, $q=1.6\times 10^{-19}\ C$

magnetic field strength, $B=0.36\ T$

we've mass of proton, $m=1.67\times 10^{-27}\ kg$

a)

From the equivalence of magnetic force and the centripetal force on the proton:

$F_B=F_C$

$q.v.B=\frac{m.v^2}{r}$

$q.B=\frac{m.v}{r}$

where:

v = speed of the proton

$(1.6\times 10^{-19})\times 0.36=\frac{1.67\times 10^{-27}\times v}{0.2}$

$v=6.898\times 10^{6}\ m.s^{-1}$ is the speed of each proton

b)

Now the centripetal force on each proton:

$F_c=m.\frac{v^2}{r}$

$F_c=1.67\times 10^{-27}\times \frac{(6.898\times 10^6)^2}{0.2}$

$F_c=3.97\times 10^{-13}\ N$