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.
Determine the force on Q at the instant it is released.
Describe the motion of Q after release.
On the diagram draw an arrow to show the direction of the magnetic field at Q due to wire X alone.
Determine the magnitude and direction of the resultant magnetic field at Q.
Markscheme
«» 18 «V» ✓
✓
«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.
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 ✓
arrow of any length as shown ✓
«using components or Pythagoras to get» B = 21 «mT» ✓
directed «horizontally» to the right ✓
If no unit seen, assume mT.