Answer:
Part a)
[tex]f = 1.46 \times 10^7 Hz[/tex]
Part b)
[tex]KE = 3.87 \times 10^{-12} J[/tex]
Explanation:
Part a)
As we know that radius of circular path of a charge moving in constant magnetic field is given as
[tex]R = \frac{mv}{qB}[/tex]
now we have
[tex]v = \frac{qBR}{m}[/tex]
now the frequency of oscillator is given as
[tex]f = \frac{v}{2\pi R}[/tex]
[tex]f = \frac{qB}{2\pi m}[/tex]
[tex]f = \frac{(1.6 \times 10^{-19})(0.960)}{2\pi(1.67\times 10^{-27})}[/tex]
[tex]f = 1.46 \times 10^7 Hz[/tex]
PART b)
now for kinetic energy of proton we will have
[tex]KE = \frac{1}{2}mv^2[/tex]
[tex]KE = \frac{1}{2}m(\frac{qBR}{m})^2[/tex]
[tex]KE = \frac{q^2B^2R^2}{2m}[/tex]
[tex]KE = \frac{(1.6 \times 10^{-19})^2(0.960)^2(0.740)^2}{2(1.67\times 10^{-27})}[/tex]
[tex]KE = 3.87 \times 10^{-12} J[/tex]
