Question

A proton moves in a circular orbit of radius 65 cm perpendicular to a uniform magneticfield of magnitude 0. 75 t

Answer 1

The speed of the proton is $\rm 4.7 \times 10^7 \ meter \ per \ second$.

What is the speed of the proton?

For the speed of a proton that has been accelerated from rest, we would apply the work-energy theorem:

A proton moves in a circular orbit with a radius of 65 cm that is perpendicular to a uniform magnetic field of magnitude 0.75 T.

The speed of the proton is given by the following formula;

$\rm V=\dfrac{2\pi R}{T}\\\\V=\dfrac{2 \times 3.14 \times 0.65m }{87.4ns}\\\\V =4.7 \times 10^7 \ meter \ per \ second$

Hence, the speed of the proton is $\rm 4.7 \times 10^7 \ meter \ per \ second$.

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