Answer:
Force on the proton will be [tex].73\times 10^{-14}N[/tex]
Explanation:
We have given speed of proton is 20.3% of speed of light
Speed of light [tex]c=3\times 10^8m/sec[/tex]
So speed of proton [tex]v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec[/tex]
Magnetic field B = 0.00629 T
Charge on proton [tex]q=1.6\times 10^{-16}C[/tex]
Angle between velocity and magnetic field [tex]\Theta =137^{\circ}[/tex]
Force on the proton is equal to [tex]F=qvBsin\Theta =1.6\times 10^{-19}\times 6.9\times 10^7\times 0.00629\times sin(137^{\circ})=4.73\times 10^{-14}N[/tex]
