Question

A proton moves through a magnetic field at 20.3 % of the speed of light. At a location where the field has a magnitude of 0.00629 T and the proton's velocity makes an angle of 137 ∘ with the field, what is the magnitude of the magnetic force acting on the proton?

Answer 1

Force on the proton will be
`.73* 10^(-14)N`

Step-by-step explanation:

We have given speed of proton is 20.3% of speed of light

Speed of light
`c=3* 10^8m/sec`

So speed of proton
`v=(3* 10^8* 23)/(100)=6.9* 10^7m/sec`

Magnetic field B = 0.00629 T

Charge on proton
`q=1.6* 10^(-16)C`

Angle between velocity and magnetic field
`\Theta =137^(\circ)`

Force on the proton is equal to
`F=qvBsin\Theta =1.6* 10^(-19)* 6.9* 10^7* 0.00629* sin(137^(\circ))=4.73* 10^(-14)N`