DP Physics (last assessment 2024)

Test builder »

Question 19M.2.HL.TZ2.5

Select a Test
Date May 2019 Marks available [Maximum mark: 7] Reference code 19M.2.HL.TZ2.5
Level HL Paper 2 Time zone TZ2
Command term Calculate, Determine, Label Question number 5 Adapted from N/A
5.
[Maximum mark: 7]
19M.2.HL.TZ2.5

A proton moves along a circular path in a region of a uniform magnetic field. The magnetic field is directed into the plane of the page.

(ai)

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

[1]

Markscheme

F towards centre ✔

Examiners report

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.

(aii)

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

[1]

Markscheme

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

 

Examiners report

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.

The speed of the proton is 2.16 × 106 m s-1 and the magnetic field strength is 0.042 T.

(bi)

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

[3]

Markscheme

« q v B = m v 2 R ⇒ » R = m v q B / 1.673 × 10 − 27 × 2.16 × 10 6 1.60 × 10 − 19 × 0.042  ✔

R = 0.538«m»  ✔

R = 0.54«m»   ✔

Examiners report

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.

(bii)

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

[2]

Markscheme

T = 2 π R v / 2 π × 0.54 2.16 × 10 6  ✔

T = 1.6 × 10 − 6 «s»   ✔

Examiners report

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

Syllabus sections