Oxford Revise AQA A Level Physics | Chapter P17 answers
P17: Magnetic fields
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Question |
Answers |
Extra information |
Mark |
AO |
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01.1 |
Reading will increase because the magnet will experience a downwards force/an equal and opposite force from the current (according to Newton’s third law) |
1 |
3.4.1.5 |
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01.2 |
Max 3 from:
Accuracy:
Safety:
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Full marks only if safety/accuracy point included |
max 3 |
3.7.5.1 |
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01.3 |
Evidence of large triangle – or clear data points taken from graph \(\begin{array}{l} \textrm{e.g. }\frac{2.0 – 0.6}{5.4 – 2.6} = 0.5 \pm 0.1\textrm{ g A}^{-1}\\ \textrm{or } 5.0 \pm 0.1 \times 10^{-4}\textrm{ kg A}^{-1}\\ \textrm{kg A}^{-1}\\ \end{array} \) |
Allow either 1 mark for correct units |
1 |
MS3.4 |
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01.4 |
\( \begin{array}{l} \textrm{Use of }F = BIl\textrm{ or }F = mg\\ mg = BIl\\ m = \frac{Bl}{g} \times I\\ \textrm{gradient } = \frac{Bl}{g}\\ B =\textrm{ gradient }\times \frac{g}{l} = (5.0 \pm 0.1 \times 10^{-4}) \times \frac{9.81}{0.05} = 0.098\textrm{ T}\\ \end{array} \) |
possible follow through from value of gradient in 01.3 |
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3.7.5.1 |
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02.1 |
ΔW = ΔVQ = eV since e is charge on proton if this happens n times energy gained is neV |
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3.7.3.3 |
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02.2 |
B field is into the page/perpendicular and into |
1 |
3.7.5.1 |
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02.3 |
Spiral path shown – exiting the cyclotron at some point |
1 |
3.7.5.2 |
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02.4 |
Electric fields:
Magnetic fields:
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3.7.3.2 |
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02.5 |
\( \begin{array}{l} F = \frac{mv^2}{r}\textrm{ or }F = Bqv\\ \frac{mv^2}{r} = Bqv\\ \frac{mv}{r} = Bq\\ v = \frac{2\pi r}{T}\textrm{ and }T = \frac{1}{f}\textrm{ so }v = 2 \pi rf\\ \frac{m \times 2\pi rf}{r} = Bq\\ m2\pi f = Bq\\ f = \frac{Bq}{2\pi m}\\ \end{array} \) |
Rearranging and cancelling must be clear in method 1 mark for each of:
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1 |
3.6.1.1 |
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02.6 |
Cyclotron frequency will decrease as the acceleration will be less (\(a = \frac{F}{m}\)) so velocity and hence period will increase |
1 |
3.5.1.4 |
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03.1 |
Electric field provides force to the right Force due to E field = qE Magnetic field provides force to the left Force due to B field = Bqv When forces are equal the ion can enter/ion undeflected qE = Bqv \(v = \frac{E}{B}\) |
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3.7.3.2 |
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03.2 |
\( \begin{array}{l} V = \frac{E}{B}\\ E = vB = 0.1 \times 4.2 \times 10^{5}\\ E = \frac{V}{d}\\ d = \frac{V}{E} = \frac{400}{0.1 \times 4.2 \times 10^5}\\ d = 0.0095\textrm{ m (0.01 m)}\\ \end{array} \) |
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3.7.3.2 |
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03.3 |
\( \begin{array}{l} F = \frac{mv^2}{r} \textrm{ or } F = Bqv\\ \frac{mv^2}{r} = Bqv\\ \frac{mv}{r} = Bq\\ r = \frac{mv}{Bq}\\ \end{array} \) |
1 |
3.6.1.1 |
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03.4 |
\( \begin{array}{l} r = \frac{mv}{Bq}\\ \Delta r = \frac{\left( 43.9 – 39.9 \right) \times 1.661 \times 10^{-27} \textrm{ kg } \times 4.2 \times 10^5}{1.1 \times 1.6 \times 10^{-19}}\\ \Delta r = 0.016\textrm{ m}\\ \end{array} \) |
1 |
3.8.1.6 |
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04.1 |
Max 2 from:
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1 |
3.7.5.4 |
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04.2 |
Angle between search coil and magnetic field:
E.m.f induced from the oscilloscope screen:
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PS4.1 |
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04.3 |
Graph of e.m.f versus cos \(\theta\) with suitable line of best fit Axes labelled with units Suitable scales chosen (data should be at least half graph paper) |
If e.m.f versus \(\theta\) plotted lose one mark |
1 |
3.7.5.3 |
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04.4 |
Need to determine intercept to check relationship – \(\begin{array}{l} \textrm{gradient } = \frac{155 – 128}{0.97 – 0.79} = 150 \pm 4\\ \textrm{Intercept}: y = mx + c\\ 155 – (150 \times 0.97) = c\\ c = 9.5\\ \end{array} \) Conclusion: this does prove the relationship as this is a straight line through the origin but there must be a systematic error OR This does not prove the relationship as there is not a straight line through the origin |
If \(\theta\) graph drawn allow one mark for stating it is cosine graph Allow either conclusion with justification. i.e. what they are looking for |
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MS3.3 |
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05.1 |
\( \begin{array}{l} \textrm{Area } = \pi r^2 = \pi \times 0.9 \times 10^{-2}\\ \phi = BA = 5.0 \times 10^{-3} \times (\pi \times 0.9 \times 10^{-2})^2\\ \phi = 1.3 \times 10^{-6}\\ \end{array} \) Wb or webers |
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3.7.5.3 |
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05.2 |
\begin{array}{l} \phi = BA\ \cos\theta = 5.0 \times 10^{-3} \times (\pi \times 0.9 \times 10^{-2})^{2}\ \cos \ 40\\ \phi = 9.7 \times 10^{-7}\textrm{ (Wb)}\\ \end{array} \) |
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3.7.5.3 |
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05.3 |
\( \begin{array}{l} \varepsilon = N \frac{\Delta \phi}{\Delta t}\\ \varepsilon = 5000 \times \frac{1.3 \times 10^{-6} – 9.7 \times 10^{-7}}{0.2}\\ \varepsilon = 0.0083\textrm{ V}\\ \end{array} \) |
e.c.f from 05.2 ignore minus sign |
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3.7.5.4 |
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05.4 |
A large e.m.f would be induced in the coil (larger than in 05.3) Rapid change in magnetic flux linkage \(\varepsilon = N \frac{\Delta \phi}{\Delta t}\) |
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3.7.5.4 |
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06.1 |
F = Bqv Electrons move to the right/towards Y/away from X Fleming’s left hand rule |
2nd mark for direction and explanation |
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3.7.5.2 |
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06.2 |
On Figure 8 left-hand side marked as positive and right-hand side marked as negative |
Allow an answer consistent with their answer to 06.1, i.e. if they think electrons move to left allow reverse labels |
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3.7.3.2 |
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06.3 |
Force due to magnetic field = force due to electric field \(\begin{array}{l} Bqv = Eq\\ \textrm{Since uniform field }E = \frac{V}{d}\\ Bqv = \frac{Vq}{d}\\ Bvd = V\\ \end{array} \) |
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3.7.5.2 |
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06.4 |
\( \begin{array}{l} V_H = Bvd \textrm{ and } v = \frac{I}{nAe}\\ V_{\textrm{H}}\ = \frac{Bdl}{nAe}\\ V_{\rm{H}}\ = \frac{Bdl}{ndte} = \frac{BI}{nte}\\ \end{array} \) |
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3.7.5.2 |
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06.5 |
\( \begin{array}{l} V_H = \frac{BI}{nte}\\ \textrm{all other values constant}\\ V_{\rm{H}} \propto \frac{1}{n}\\ \end{array} \) So VH greater, easier to detect magnetic flux density. |
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3.7.5.2 |
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07.1 |
micrometer/digital calliper measuring several places along length and finding mean |
1 |
Ate |
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07.2 |
\( \begin{array}{l} \rho = \frac{M}{V}\\ M = \rho \times V = 2700 \times 15 \times 10^{-3} \times 50 \times 10^{-3} \times 0.02 \times 10^{-3}\\ M = 4.05 \times 10^{-5}\textrm{ kg}\\ \end{array} \) |
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3.4.2.1 |
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07.3 |
\( \begin{array}{l} \textrm{use of }F = BIl\textrm{ or }W = mg\\ I = \frac{mg}{Bl} = \frac{4.05 \times 10^{-5}\textrm{ kg} \times 9.81}{0.03 \times 0.05}\\ I = 0.26\textrm{ A}\\ \end{array} \) |
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3.7.5.1 |
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07.4 |
Diagram showing field perpendicular to current Direction and current such that foil feels upwards force |
1 |
3.7.5.1 |
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08.1 |
Crosses drawn on the diagram – uniformly placed – at least 4 |
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3.7.5.2 |
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08.2 |
Ion feels a resultant force perpendicular to direction of motion Provides centripetal force |
1 |
3.6.1.1 |
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08.3 |
\( \begin{array}{l} F = \frac{mv^2}{r}\textrm{ or }F = Bqv\\ \frac{mv^2}{r} = Bqv\\ \frac{mv}{r} = Bq\\ r = \frac{mv}{Bq}\\ \end{array} \) |
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3.6.1.1 |
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08.4 |
Circle starting at P but then with a greater radius arriving at a point further to the right than R |
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3.7.5.2 |
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08.5 |
\( \begin{array}{l} \frac{r_1}{m_1} = \frac{r_2}{m_2}\\ r_2 – r_1 = 0.2\textrm{ mm}\\ r_2 – r_2 \frac{m_1}{m_2} = 0.2\\ r_2 (1 -\frac{m_1}{m_2}) = 0.2\\ r_2 \left( 1 – \frac{10.012937\textrm{ u}}{11.009305\textrm{ u}} \right) = 0.2\\ r_2 = 2.2\textrm{ m}\\ d = 2 \times 2.2 = 4.4\textrm{ m}\\ \end{array} \) |
Award for appreciation r proportional to m |
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3.7.5.2 |
Skills box answers
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Question |
Answer |
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1 |
Draw a graph using the data. Note that the mass is given in grams and needs to be converted to kg. The gradient of the graph is 4.0 × 10−4 kg A−1. This gives a magnetic field of 0.08 T. |
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2 |
The variable resistor is used to limit the current through the wire and to prevent it overheating. |
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3 |
The force will be reduced because \(F = BIL\sin \theta \) and now \(\theta\) is less than 90° |