DP Physics (first assessment 2025)

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Question 19N.2.SL.TZ0.4

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Date November 2019 Marks available [Maximum mark: 8] Reference code 19N.2.SL.TZ0.4
Level SL Paper 2 Time zone TZ0
Command term Calculate, Explain, Show that Question number 4 Adapted from N/A
4.
[Maximum mark: 8]
19N.2.SL.TZ0.4

A proton is moving in a region of uniform magnetic field. The magnetic field is directed into the plane of the paper. The arrow shows the velocity of the proton at one instant and the dotted circle gives the path followed by the proton.

The speed of the proton is 2.0 × 106 m s–1 and the magnetic field strength B is 0.35 T.

(a)

Explain why the path of the proton is a circle.

[2]

Markscheme

magnetic force is to the left «at the instant shown»
OR
explains a rule to determine the direction of the magnetic force ✔

force is perpendicular to velocity/«direction of» motion
OR
force is constant in magnitude ✔

force is centripetal/towards the centre ✔

NOTE: Accept reference to acceleration instead of force

(b(i))

Show that the radius of the path is about 6 cm.

[2]

Markscheme

qvB=mv2R✔

R=1.67×10-27×2.0×1061.6×10-19×0.35 OR 0.060 « m »

NOTE: Award MP2 for full replacement or correct answer to at least 2 significant figures

(b(ii))

Calculate the time for one complete revolution.

[2]

Markscheme

T=2πRv✔

T=«2π×0.062.0×106=1.9×10-7 « s » ✔

NOTE: Award [2] for bald correct answer

(c)

Explain why the kinetic energy of the proton is constant.

[2]

Markscheme

ALTERNATIVE 1
work done by force is change in kinetic energy ✔
work done is zero/force perpendicular to velocity ✔

NOTE: Award [2] for a reference to work done is zero hence Ek remains constant

 

ALTERNATIVE 2
proton moves at constant speed ✔
kinetic energy depends on speed ✔

NOTE: Accept mention of speed or velocity indistinctly in MP2

 

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