A cyclotron has two D-shaped regions where the magnetic flux density is constant - AQA - A-Level Physics - Question 3 - 2017 - Paper 2

To derive the expression, we start with the physics of circular motion. The centripetal force required to keep the proton in a circular path is provided by the magnetic force:

$F = qvB$

where:

• $F$ is the centripetal force,
• $q$ is the charge of the proton,
• $v$ is the speed of the proton,
• $B$ is the magnetic flux density.

For circular motion, we have:

$F = \frac{mv^2}{r}$

Equating the two forces gives:

$qvB = \frac{mv^2}{r}$

From this, rearranging to find $t$, the time for one full semi-circular path (half a full circle) yields:

$t = \frac{2\pi r}{v}$

Since $v$ can be expressed from the original equations, we find that $t$ ultimately does not depend on $r$.

To calculate the maximum speed, we use the given values for the magnetic flux density and the radius. The expression for speed derived is:

$v = \frac{qBr}{m}$

Using the following values:

• $q = 1.6 \times 10^{-19}\, C$ (charge of the proton),
• $B = 0.44\, T$,
• $r = 0.55\, m$, and,
• $m = 1.67 \times 10^{-27}\, kg$ (mass of the proton).

Inserting these into the equation gives:

$v = \frac{(1.6 \times 10^{-19}) (0.44) (0.55)}{(1.67 \times 10^{-27})}\approx 2.43 \times 10^{7}\, m/s$

Thus, the maximum speed of the proton when it leaves the cyclotron is approximately $2.43 \times 10^{7}\, m/s$.