Label with arrows on the diagram the magnetic force F on the proton.

Label with arrows on the diagram the velocity vector v of the proton.

For this proton, determine, in m, the radius of the circular path. Give your answer to an appropriate number of significant figures.

For this proton, calculate, in s, the time for one full revolution.

F towards centre ✔

v tangent to circle and in the direction shown in the diagram ✔

«$qvB=\frac{m{v}^2}{R}\Rightarrow »R=\frac{mv}{qB}/\frac{1.673\times {10}^{-27}\times 2.16\times {10}^6}{1.60\times {10}^{-19}\times 0.042}$  ✔

R = 0.538«m»  ✔

R = 0.54«m»   ✔

$T=\frac{2\pi R}{v}/\frac{2\pi \times 0.54}{2.16\times {10}^6}$  ✔

$T=1.6\times {10}^{-6}$«s»   ✔

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

Examiners were requested to be lenient here and as a result most candidates scored both marks. Had we insisted on e.g. straight lines drawn with a ruler or a force arrow passing exactly through the centre of the circle very few marks would have been scored. For those who didn’t know which way the arrows were supposed to be the common guesses were to the left and up the page. Some candidates neglected to label the arrows.

This was generally well answered although usually to 3 sf. Common mistakes were to substitute 0.042 for F and 1 for q. Also some candidates tried to answer in terms of electric fields.

This was well answered with many candidates scoring ECF from the previous part.