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Question 19M.2.SL.TZ2.b

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Date May 2019 Marks available [Maximum mark: 3] Reference code 19M.2.SL.TZ2.b
Level SL Paper 2 Time zone TZ2
Command term Determine Question number b Adapted from N/A
b.
[Maximum mark: 3]
19M.2.SL.TZ2.b

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.

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

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