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fredd [130]
3 years ago
6

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 0.0062

9 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?
Physics
1 answer:
galina1969 [7]3 years ago
4 0

Answer:

Force on the proton will be .73\times 10^{-14}N

Explanation:

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

Speed of light c=3\times 10^8m/sec

So speed of proton v=\frac{3\times 10^8\times 23}{100}=6.9\times 10^7m/sec

Magnetic field B = 0.00629 T

Charge on proton q=1.6\times 10^{-16}C

Angle between velocity and magnetic field \Theta =137^{\circ}

Force on the proton is equal to 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

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