Edexcel iGCSE Physics (4PH1) 6.2 Magnetism Exam Style Question Paper 2B - New Syllabus
Question
The diagram shows a machine that can be used to measure the speed of fast-moving protons.
(a) At the start, the proton is attracted towards a negatively charged plate.
(i) Give a reason why the proton is attracted to the negatively charged plate.
(ii) The proton accelerates at \(1.90\times10^{11}\,\mathrm{m\,s^{-2}}\) from rest to a speed of \(1.38\times10^5\,\mathrm{m\,s^{-1}}\). Show that the time taken for this acceleration is about \(7\times10^{-7}\,\mathrm{s}\).
(b) The proton passes through a hole in the negatively charged plate and enters an area where there is a magnetic field. The magnetic field exerts a force on the proton, as shown in the diagram. This force causes the proton to follow a circular path without changing speed.
(i) Give the direction of the magnetic field.
(ii) Suggest how increasing the strength of the magnetic field will affect the proton when it is moving in the magnetic field.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 1.6: Acceleration — part (a)(ii)
• 6.11P: Force on a Charged Particle in a Magnetic Field — parts (b)(i) and (b)(ii)
▶️ Answer/Explanation
(a)(i) Attraction between charges [1 mark]
The proton has a positive charge and the plate has a negative charge. Opposite charges attract.
Answer: Opposite charges attract.
(a)(ii) Time taken for acceleration [3 marks]
Use the equation:
\(a=\dfrac{v-u}{t}\)
The proton starts from rest, so:
\(u=0\,\mathrm{m\,s^{-1}}\)
Therefore:
\(1.90\times10^{11}=\dfrac{1.38\times10^5}{t}\)
Rearranging:
\(t=\dfrac{1.38\times10^5}{1.90\times10^{11}}\)
\(t=7.26\times10^{-7}\,\mathrm{s}\)
Therefore:
\(\boxed{t\approx7\times10^{-7}\,\mathrm{s}}\)
(b)(i) Direction of the magnetic field [1 mark]
The direction of the magnetic field is:
\(\boxed{\text{into the page}}\)
This direction is determined using the force on a moving positive charge and the direction of its motion shown in the diagram.
(b)(ii) Effect of increasing magnetic field strength [2 marks]
For a charged particle moving perpendicular to a magnetic field:
\(F=qvB\)
- Increasing the magnetic field strength \(B\) increases the magnetic force on the proton.
- This increases the centripetal acceleration, causing the proton to move in a tighter circular path.
- The proton would therefore follow a circle with a smaller radius.
The speed does not change because the magnetic force acts perpendicular to the proton’s direction of motion.
Question
This question is about electromagnets.
(a) Describe the construction of a simple electromagnet that is producing a magnetic field. You may draw a diagram to help your answer.
(b) A proton moves through a uniform magnetic field produced by a strong electromagnet. The shaded area in the diagram represents the magnetic field. The initial velocity, \(v\), of the proton is also shown.
(i) Use the left-hand rule to determine the direction of the force acting on the proton.
(ii) Explain how the force on the proton changes as the proton moves through the magnetic field. You may add to the diagram to help your answer.
(iii) Suggest why the velocity of the proton changes.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 6.9P: Construction of Electromagnets — part (a)
• 6.11P: Force on a Charged Particle in a Magnetic Field — parts (b)(i)–(iii)
▶️ Answer/Explanation
(a) Construction of an electromagnet [3 marks]
- Wind a coil or solenoid of insulated wire around a core.
- Pass a direct current through the coil.
- Use a soft iron core to strengthen the magnetic field.
When current flows through the coil, a magnetic field is produced around the wire. The soft iron core becomes magnetised and increases the strength of the electromagnet.
(b)(i) Direction of force [1 mark]
Using the left-hand rule, the force on the proton acts:
\(\boxed{\text{upwards, towards the top of the page}}\)
(b)(ii) Change in force [4 marks]
As the proton moves through the uniform magnetic field:
- The magnetic force remains at right angles to the velocity of the proton.
- The force changes the direction of motion of the proton.
- As the proton’s direction changes, the direction of the force also changes because the force depends on the direction of motion of the charged particle.
- The magnitude of the force remains constant while the proton moves through the uniform field, provided its speed remains constant.
The proton therefore follows a curved, approximately circular path while it remains within the uniform magnetic field. The magnetic force acts as a centripetal force.
(b)(iii) Why the velocity changes [1 mark]
Velocity is a vector quantity, so it changes whenever either its magnitude or its direction changes.
The magnetic force changes the direction of the proton’s motion, so the velocity changes even though the speed remains constant.
Question
This question is about electromagnetism.
(a) Diagram 1 shows the magnetic field around a straight section of copper wire.
Explain why the copper wire has the magnetic field shown in the diagram.
(b) A student investigates how the strength of an electromagnet varies with the current in the electromagnet. The diagram shows their apparatus.
This is the student’s method.
- switch on the electromagnet at its maximum current
- place a load of \(100\,\mathrm{g}\) so that it is held above the floor by the electromagnet
- slowly reduce the current in the electromagnet until the load falls from the electromagnet
- record the current at which the load falls
- record the current at which the same load falls two more times
Repeat the method for loads of different masses.
(i) Suggest a suitable safety precaution for the student’s investigation.
(ii) The table shows the student’s results.
Calculate the mean current when the mass of the load was \(600\,\mathrm{g}\). Give your answer to a suitable number of significant figures.
(iii) On the grid, plot a graph of the mean current against the mass of the load. The scale for the mass axis has been done for you.
(iv) Draw the line of best fit.
(v) The student predicts that a load of \(1.0\,\mathrm{kg}\) will fall when the current in the electromagnet is \(3.0\,\mathrm{A}\). Comment on the student’s prediction.
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 6.9P: Construction of Electromagnets — parts (b), (i), (ii), (iii), (iv) and (v)
▶️ Answer/Explanation
(a) Magnetic field around a current-carrying wire [2 marks]
- There must be a current flowing through the copper wire.
- The current must be flowing to the right.
A current-carrying conductor produces a magnetic field around it. The direction of the magnetic field depends on the direction of the current.
(b)(i) Safety precaution [1 mark]
One suitable precaution is to keep hands and feet away from the load so that they are not hit when the load falls.
Other suitable precautions include taking care to avoid the heating effect of the current or protecting the floor from damage caused by falling loads.
(b)(ii) Mean current for \(600\,\mathrm{g}\) [2 marks]
Calculate the mean of the three current readings for the \(600\,\mathrm{g}\) load:
\(\mathrm{mean\ current}=\dfrac{\mathrm{sum\ of\ readings}}{3}\)
The calculated mean is:
\(\mathrm{mean\ current}=1.833\ldots\,\mathrm{A}\)
To \(3\) significant figures:
\(\boxed{1.83\,\mathrm{A}}\)
(b)(iii) Plotting the graph [3 marks]
- Use a sensible, continuous scale on the current axis so that the plotted data covers a significant portion of the grid.
- Label the vertical axis as current, \(I/\mathrm{A}\).
- Plot all the data points accurately using the calculated mean currents.
(b)(iv) Line of best fit [1 mark]
Draw a straight line of best fit through the data, with approximately equal numbers of points distributed on either side of the line.
(b)(v) Comment on the prediction [3 marks]
- \(1.0\,\mathrm{kg}=1000\,\mathrm{g}\).
- The trend in the results indicates that the current required increases as the mass of the load increases.
- The data suggests that current is approximately directly proportional to mass, so extrapolation may give a value close to \(3.0\,\mathrm{A}\).
- However, \(1000\,\mathrm{g}\) is outside the range of the masses tested, so the prediction involves extrapolation.
- The pattern observed within the measured range may not continue beyond the range of data collected.
Therefore, the prediction of \(3.0\,\mathrm{A}\) may be reasonable based on the trend, but it is less reliable because \(1.0\,\mathrm{kg}\) is outside the experimental range.
Question
The diagram shows the inside of an oscilloscope. Electrons leave the electron supply and accelerate towards the screen. The stream of electrons hits the screen.
(a) The grid is connected to the positive terminal of a high voltage power supply.
(i) Explain, in terms of the movement of particles, how the grid has become positively charged.
(ii) State the formula linking kinetic energy, mass and speed.
(iii) Calculate the speed of an electron when it has \(1.3\times10^{-15}\,\mathrm{J}\) of kinetic energy. The mass of an electron is \(9.1\times10^{-31}\,\mathrm{kg}\).
(b) The stream of electrons spreads out as it travels towards the screen. Explain why the electrons in the stream move apart from each other.
(c) The oscilloscope can cause the electrons to move in a circle. This produces a circular pattern on the screen.
(i) Use the scale to determine the radius of the circle.
(ii) The time taken for the electrons to complete one orbit of the circle is \(24\,\mathrm{ms}\). Calculate the orbital speed of the electrons. Use the formula
\(\displaystyle \mathrm{orbital\ speed}=\frac{2\pi\times\mathrm{orbital\ radius}}{\mathrm{time\ period}}\)
Syllabus Topic Codes (Edexcel International GCSE Physics 4PH1):
• 4.14: Kinetic Energy — parts (a)(ii)–(iii)
• 2.25P: Forces Between Electric Charges — part (b)
• 6.11P: Force on a Charged Particle in a Magnetic Field — part (c)
▶️ Answer/Explanation
(a)(i) Positive charging [2 marks]
The grid becomes positively charged because it loses electrons.
Electrons are transferred away from the grid, leaving it with more positive charge than negative charge.
(a)(ii) Kinetic energy formula [1 mark]
The formula linking kinetic energy, mass and speed is:
\(\boxed{E_\mathrm{k}=\dfrac{1}{2}mv^2}\)
(a)(iii) Speed of the electron [3 marks]
Use:
\(E_\mathrm{k}=\dfrac{1}{2}mv^2\)
Substitute the values:
\(1.3\times10^{-15}=\dfrac{1}{2}\times9.1\times10^{-31}\times v^2\)
Rearranging:
\(\displaystyle v^2=\frac{2(1.3\times10^{-15})}{9.1\times10^{-31}}\)
\(v^2\approx2.86\times10^{15}\)
Therefore:
\(v=\sqrt{2.86\times10^{15}}\)
\(v\approx5.35\times10^7\,\mathrm{m\,s^{-1}}\)
\(\boxed{v=5.3\times10^7\,\mathrm{m\,s^{-1}}}\)
(b) Repulsion between electrons [2 marks]
All the electrons in the stream have the same type of charge, so they are like charges.
Like charges repel each other. Therefore, the electrons exert repulsive forces on one another and move apart as they travel towards the screen.
(c)(i) Radius of the circle [1 mark]
From the scale on the diagram, the radius of the circular pattern is:
\(\boxed{r=2.5\,\mathrm{cm}}\)
(c)(ii) Orbital speed [3 marks]
Convert the quantities into SI units:
\(r=2.5\,\mathrm{cm}=2.5\times10^{-2}\,\mathrm{m}\)
\(T=24\,\mathrm{ms}=24\times10^{-3}\,\mathrm{s}\)
Use:
\(\displaystyle v=\frac{2\pi r}{T}\)
\(\displaystyle v=\frac{2\pi(2.5\times10^{-2})}{24\times10^{-3}}\)
\(v\approx6.54\,\mathrm{m\,s^{-1}}\)
Therefore:
\(\boxed{v=6.5\,\mathrm{m\,s^{-1}}}\)