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Date November 2021 Marks available 1 Reference code 21N.2.SL.TZ0.4
Level Standard level Paper Paper 2 Time zone 0 - no time zone
Command term Calculate Question number 4 Adapted from N/A

Question

A charged particle, P, of charge +68 μC is fixed in space. A second particle, Q, of charge +0.25 μC is held at a distance of 48 cm from P and is then released.

The diagram shows two parallel wires X and Y that carry equal currents into the page.

Point Q is equidistant from the two wires. The magnetic field at Q due to wire X alone is 15 mT.

The work done to move a particle of charge 0.25 μC from one point in an electric field to another is 4.5 μJ. Calculate the magnitude of the potential difference between the two points.

[1]
a.

Determine the force on Q at the instant it is released.

[2]
b.i.

Describe the motion of Q after release.

[2]
b.ii.

On the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone.

[1]
c.i.

Determine the magnitude and direction of the resultant magnetic field at Q.

[2]
c.ii.

Markscheme

«V=4.50.25=» 18 «V» ✓

a.

F=8.99×109×68×10-6×0.25×10-60.482 ✓

F=0.66 «N» ✓

 

Award [2] marks for a bald correct answer.

Allow symbolic k in substitutions for MP1.

Do not allow ECF from incorrect or not squared distance.

b.i.

Q moves to the right/away from P «along a straight line»

OR

Q is repelled from P ✓


with increasing speed/Q accelerates ✓

acceleration decreases ✓

b.ii.

 

arrow of any length as shown ✓

c.i.

«using components or Pythagoras to get» B = 21 «mT» ✓

directed «horizontally» to the right ✓

 

If no unit seen, assume mT.

c.ii.

Examiners report

[N/A]
a.
[N/A]
b.i.
[N/A]
b.ii.
[N/A]
c.i.
[N/A]
c.ii.

Syllabus sections

Core » Topic 5: Electricity and magnetism » 5.1 – Electric fields
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