kinetic energy

[logistic_guy]

A \(\displaystyle 5.0\)-\(\displaystyle \text{MeV}\) (kinetic energy) proton enters a \(\displaystyle 0.20\)-\(\displaystyle \text{T}\) field, in a plane perpendicular to the field. What is the radius of its path?

 

[khansaheb]

A \(\displaystyle 5.0\)-\(\displaystyle \text{MeV}\) (kinetic energy) proton enters a \(\displaystyle 0.20\)-\(\displaystyle \text{T}\) field, in a plane perpendicular to the field. What is the radius of its path?

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[logistic_guy]

We use the magnetic force formula to solve this problem.

\(\displaystyle F = evB = ma\)

When the acceleration is centripetal \(\displaystyle a = \frac{v^2}{r}\).

Then

\(\displaystyle evB = m\frac{v^2}{r}\)

\(\displaystyle eB = m\frac{v}{r}\)

Kinetic energy formula is:

\(\displaystyle K = \frac{1}{2}mv^2\)

Or

\(\displaystyle 2mK = m^2v^2\)

Then,

\(\displaystyle eB = m\frac{v}{r}\)

\(\displaystyle eBr = mv\)

\(\displaystyle e^2B^2r^2 = m^2v^2 = 2mK\)

\(\displaystyle r = \frac{\sqrt{2mK}}{eB}\)

Don't forget to convert \(\displaystyle \text{eV}\) to \(\displaystyle \text{J}\).

\(\displaystyle r = \frac{\sqrt{2(1.67 \times 10^{-27})(5 \times 10^{6})(1.6 \times 10^{-19})}}{(1.6 \times 10^{-19})(0.2)} = \textcolor{blue}{1.62 \ \text{m}}\)