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 use the centripetal motion equation where the magnetic force provides the necessary centripetal force:

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

The magnetic force experienced by the proton is given by:

$F = BQv$

Setting the two forces equal gives:

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

Rearranging leads to:

$t = \frac{\text{distance}}{\text{speed}} = \frac{\pi r}{v}$

From earlier, substituting for $v$:

$v = \frac{BQ}{m} \cdot r$

Thus:

$t = \frac{\pi r}{\frac{BQ}{m} \cdot r}$

Which simplifies to:

$t = \frac{\pi m}{BQ}$

This expression shows that the time taken is independent of the radius.

Using the earlier derived expression for speed:

$v = \frac{BQ}{m} \cdot r$

Substituting the known values:

• $B = 0.44 \, T$
• $Q = 1.6 \times 10^{-19} \, C$
• $m = 1.67 \times 10^{-27} \, kg$
• $r = 0.55 \, m$

Calculating:

$v = \frac{(0.44)(1.6 \times 10^{-19})}{1.67 \times 10^{-27}} \cdot 0.55$

Performing the calculation:

$v \approx 2.43 \times 10^7 \, m/s$

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