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A-Level · Physics

AP9 Physics: Fields (Gravitational, Electric, Magnetic)

AQA 7408 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

A navigation satellite of mass 640 kg moves in a circular orbit around the Earth at constant speed, held in orbit entirely by the Earth's gravitational attraction. Treat the Earth as a uniform sphere of mass M = 5.97 x 10²⁴ kg and radius R = 6.37 x 10⁶ m, and take G = 6.67 x 10^-11 N m² kg^-2. You are given Newton's law of gravitation, F = GMm/r², and the definition of gravitational field strength, g = F/m, where m is the mass placed in the field.

(a)Show that the gravitational field strength at a distance r from the centre of the Earth is given by g = GM/r^2.(2)

(b)The satellite orbits at a height of 2.00 x 10⁷ m above the Earth's surface. Calculate the gravitational field strength at the position of the satellite.(3)

(c)By equating the gravitational force on the satellite to the centripetal force needed for circular motion, and using v = 2*π*r/T, show that the orbital period T of the satellite is given by T = 2*π*√r³/(GM).(3)

(d)Hence calculate the orbital period of the satellite, giving your answer in hours.(3)

(Total for Question 1 is 11 marks)

A meteoroid of mass 2.50 x 10³ kg travels from deep space towards the Earth. Treat the Earth as an isolated uniform sphere (ignore the Sun and Moon), take the gravitational potential at infinity to be zero, and use G = 6.67 x 10^-11 N m² kg^-2, M(Earth) = 5.97 x 10²⁴ kg, R(Earth) = 6.37 x 10⁶ m. You are given: gravitational potential V = -GM/r, and escape velocity v = √2GM/r.

(a)Calculate the gravitational potential at the Earth's surface.(2)

(b)Calculate the minimum kinetic energy the meteoroid would need at the Earth's surface so that it could, in principle, escape to infinity (ignore atmospheric resistance).(3)

(c)Show that the escape velocity from the Earth's surface is about 1.12 x 10⁴ m/s.(3)

(d)The meteoroid actually approaches the Earth with a speed of 1.60 x 10⁴ m/s while still very far away (effectively at infinity). Using conservation of energy, and ignoring atmospheric drag, calculate its speed just above the atmosphere, at a distance R from the Earth's centre.(4)

(Total for Question 2 is 12 marks)

A student drops a strong bar magnet, from rest, down through a long vertical tube. In one trial the tube is made of a conducting metal such as copper; in a second, otherwise identical, trial the tube is made of an insulating plastic. The magnet takes noticeably longer to fall through the copper tube than through the plastic tube, even though it is dropped from the same height each time and air resistance can be ignored in both cases.

Explain, in terms of Faraday's law and Lenz's law, why the magnet falls with a smaller acceleration in the copper tube than in the plastic tube, and explain what happens to the gravitational potential energy the magnet loses as it falls through the copper tube.

Figure (to be drawn): A vertical hollow tube with a bar magnet falling through its centre, oriented with its axis vertical; the tube wall is shown as a continuous conducting sheet in one case and as an insulator in the other.

(Total for Question 3 is 6 marks)

Two small charged spheres, P and Q, are fixed 0.360 m apart in a vacuum. Sphere P carries a charge of +4.80 x 10^-9 C and sphere Q carries a charge of -7.20 x 10^-9 C. You are given 1/(4*π*epsilon0) = 8.99 x 10⁹ N m² C^-2, so that the force between two point charges is F = Q1*Q2/(4*π*epsilon0*r²) and the electric field due to a point charge is E = Q/(4*π*epsilon0*r²).

Figure (to be drawn): Two point charges P (+4.80 nC) and Q (-7.20 nC) fixed 0.360 m apart on a horizontal line, with the midpoint marked.

(a)Calculate the magnitude of the electrostatic force between P and Q, and state whether the force is attractive or repulsive.(3)

(b)Calculate the electric field strength due to P alone at the location of Q.(2)

(c)A small test charge is placed at the midpoint between P and Q, 0.180 m from each sphere. Given that the field due to P and the field due to Q both point in the same direction at this point (from P towards Q), calculate the magnitude of the resultant electric field at the midpoint.(4)

(d)State the direction of the resultant electric field at the midpoint.(1)

(Total for Question 4 is 10 marks)

A small charged oil drop of mass 9.80 x 10^-15 kg carries a charge of -3.20 x 10^-19 C. The drop is held stationary between two horizontal parallel plates separated by a distance of 5.00 x 10^-3 m, in a Millikan-type experiment. Take g = 9.81 N kg^-1.

Figure (to be drawn): Two horizontal parallel plates 5.00 x 10^-3 m apart, with a small negatively charged oil drop shown stationary midway between them.

(a)Show that the electric field strength required to hold the drop stationary is about 3.00 x 10⁵ N/C.(3)

(b)Calculate the potential difference between the plates needed to produce this field.(2)

(c)State which plate, upper or lower, must be at the higher potential, explaining your reasoning.(2)

(Total for Question 5 is 7 marks)

In the circuit shown, a capacitor C1 of capacitance 150 uF is connected in series with a parallel combination of two capacitors, C2 = 180 uF and C3 = 270 uF. The combination is connected to a 15.0 V d.c. supply and allowed to charge fully.

Figure (to be drawn): Series circuit: 15.0 V d.c. supply connected across C1 (150 uF), in series with C2 (180 uF) connected in parallel with C3 (270 uF).

(a)Calculate the combined capacitance of C2 and C3 in parallel.(1)

(b)Calculate the total capacitance of the circuit.(3)

(c)Calculate the total charge supplied by the 15.0 V source when the capacitor network is fully charged.(2)

(d)Calculate the potential difference across C1.(2)

(e)Calculate the total energy stored in the fully charged capacitor network.(3)

(Total for Question 6 is 11 marks)

REQUIRED PRACTICAL. A student carries out the required practical to investigate how the potential difference across a discharging capacitor varies with time. A capacitor of capacitance 300 uF is charged to 8.00 V then discharged through a resistor of resistance 47.0 kOhm, with a data logger recording the potential difference V across the capacitor at regular time intervals. You are given: V = V0 * exp(-t/(RC)), where V0 is the potential difference at t = 0.

Figure (to be drawn): Circuit diagram: a charged 300 uF capacitor connected through a switch to a 47.0 kOhm resistor, with a data logger (voltage sensor) connected across the capacitor to record V against t during discharge.

(a)Calculate the time constant of this circuit, stating an appropriate unit.(2)

(b)Calculate the initial charge stored on the capacitor.(2)

(c)Calculate the potential difference across the capacitor 20.0 s after the discharge begins.(3)

(d)Describe how the student could use their V against t data to determine the time constant RC graphically, without directly reading a value from the decay curve itself.(3)

(e)State one precaution the student should take to improve the accuracy of the reading taken at t = 0.(1)

(Total for Question 7 is 11 marks)

REQUIRED PRACTICAL. A student determines the magnetic flux density between the poles of a strong magnet using a current balance. A straight horizontal wire of length 9.00 x 10^-2 m is placed at right angles to the magnetic field, resting on a support connected to a top-pan balance. You are given F = B*I*L for a wire at right angles to a uniform field. Take g = 9.81 N kg^-1.

Figure (to be drawn): A horizontal wire of length 9.00 cm passes at right angles through the field between two magnadur magnets mounted on a yoke; the magnet assembly rests on a top-pan balance so that the reaction to the force on the wire changes the balance reading when current flows.

(a)With no current flowing, the balance reads 152.40 g. When a current of 3.50 A flows through the wire, the balance reading changes to 152.87 g. Calculate the magnetic force on the wire corresponding to this change in balance reading.(3)

(b)Calculate the magnetic flux density between the poles of the magnet.(2)

(c)The student repeats the experiment for a range of currents and plots a graph of force F (y-axis) against current I (x-axis). Explain how this graph gives a more reliable value of B than using a single pair of readings.(3)

(Total for Question 8 is 8 marks)

In a simple mass spectrometer, singly ionised carbon-12 ions (charge +1.60 x 10^-19 C, mass 1.99 x 10^-26 kg) are accelerated from rest through a potential difference of 2.00 x 10³ V and then enter a region of uniform magnetic flux density 0.450 T, directed at right angles to their velocity. You are given: work done accelerating the ion, q*V = (1/2)*m*v², and the condition for circular motion in the magnetic field, B*q*v = m*v²/r.

Figure (to be drawn): Ions are accelerated from rest through a potential difference between two plates, then travel in a circular arc within a region of uniform magnetic field directed into the page.

(a)Calculate the speed of the carbon-12 ions as they enter the magnetic field.(3)

(b)Calculate the radius of the circular path followed by the carbon-12 ions in the magnetic field.(3)

(c)A singly ionised carbon-13 ion (same charge, mass 2.16 x 10^-26 kg) is accelerated through the same potential difference and enters the same magnetic field. Without recalculating the speed and radius from scratch, state and explain whether the radius of its circular path would be greater than, less than, or equal to that of the carbon-12 ion.(3)

(Total for Question 9 is 9 marks)

A simple a.c. generator consists of a rectangular coil of 300 turns, each of area 1.10 x 10^-2 m², rotating at a constant angular frequency of 300 rad/s in a uniform magnetic field of flux density 0.0750 T. You are given: the EMF induced by a rotating coil, EMF = B*A*N*ω*sin(ω*t).

Figure (to be drawn): A rectangular coil of 300 turns rotating about a vertical axis within a uniform horizontal magnetic field, with slip rings connecting the coil to an external circuit.

(a)Calculate the peak EMF generated by the coil.(2)

(b)Calculate the frequency of rotation of the coil, in Hz.(2)

(c)Describe how the induced EMF varies with time over one complete rotation of the coil, and explain, in terms of the rate of change of flux linkage, why the EMF is zero at certain points in the rotation.(3)

(Total for Question 10 is 7 marks)

A step-down transformer converts a 230 V a.c. mains supply to a 12.0 V supply for a low-voltage lighting system. The primary coil has 4600 turns. You are given the ideal transformer equation Vs/Vp = Ns/Np.

(a)Calculate the number of turns needed on the secondary coil, assuming the transformer is ideal.(2)

(b)In practice this transformer is only 92.0% efficient. When the secondary current is 8.00 A, calculate the primary current, assuming Vp and Vs are unchanged.(4)

(Total for Question 11 is 6 marks)

A velocity selector consists of a region where a uniform electric field and a uniform magnetic field act at right angles to each other and to the velocity of charged particles passing through. A beam of ions passes through undeviated when the electric force and the magnetic force on each ion are equal and opposite. The electric field strength is 6.00 x 10⁴ V/m and the magnetic flux density in the selector is 0.120 T.

Figure (to be drawn): A velocity selector: ions travel horizontally through a region with a vertical electric field between two plates and a perpendicular magnetic field (into the page); only ions of one particular speed pass through undeviated to a small exit aperture, after which they enter a separate magnetic field region and follow a circular path.

(a)Show that the speed of ions that pass through the selector undeviated is given by v = E/B, and calculate this speed for the values given.(3)

(b)Ions leaving the selector then enter a separate uniform magnetic field of flux density 0.350 T, directed perpendicular to their velocity, where they follow a circular path of radius 4.20 x 10^-2 m. Calculate the specific charge (charge-to-mass ratio, q/m) of the ions.(3)

(c)Suggest why, in practice, it can be difficult to obtain a truly uniform electric field over the whole region through which the ions travel in the velocity selector, and state one consequence of this for the beam of ions leaving the selector.(2)

(Total for Question 12 is 8 marks)

Mark scheme · AP9 Physics: Fields (Gravitational, Electric, Magnetic)

Question 1

(a) B1 states g = F/m (or equivalent definition of field strength as force per unit mass)
(a) B1 substitutes F = GMm/r² and cancels m correctly to reach g = GM/r² (cao)
(a) Answer: g = GM/r²
(b) M1 correctly finds r = h + R = 2.00 x 10⁷ + 6.37 x 10⁶ = 2.637 x 10⁷ m
(b) M1 substitutes GM = 3.982 x 10¹⁴ and their r into g = GM/r²
(b) A1 g = 0.573 N kg^-1 (awrt 3 sf), allow ecf from their r
(b) Answer: g = 0.573 N kg^-1
(c) B1 equates gravitational force to centripetal force: GMm/r² = m*v²/r (or m*ω²*r)
(c) B1 substitutes v = 2*π*r/T and rearranges to GM/r² = 4*π²*r/T²
(c) B1 correctly rearranges to T² = 4*π²*r³/(GM), hence T = 2*π*√r³/(GM) (cao)
(c) Answer: T = 2*π*√r³/(GM)
(d) M1 substitutes r = 2.637 x 10⁷ m and GM = 3.982 x 10¹⁴ into T = 2*π*√r³/(GM)
(d) A1 correctly evaluates r³/(GM) = 4.61 x 10⁷ (or equivalent intermediate value), giving T = 4.26 x 10⁴ s
(d) A1 converts to T = 11.8 hours (awrt 3 sf), allow ecf
(d) Answer: T = 4.26 x 10⁴ s = 11.8 hours

Question 2

(a) M1 substitutes correctly into V = -GM/R
(a) A1 V = -6.25 x 10⁷ J kg^-1 (awrt 3 sf), correct sign and unit
(a) Answer: V = -6.25 x 10⁷ J kg^-1
(b) M1 recognises that total mechanical energy per unit mass must be at least zero for escape: KE_min = -m*V(surface)
(b) A1 substitutes m = 2.50 x 10³ kg and their V from part (a)
(b) A1 KE_min = 1.56 x 10¹¹ J (awrt 3 sf), allow ecf
(b) Answer: KE_min = 1.56 x 10¹¹ J
(c) M1 quotes and uses v = √2GM/R
(c) A1 correct substitution of GM and R
(c) A1 evaluates to v = 1.12 x 10⁴ m/s (cao), matching the value given
(c) Answer: v = 1.12 x 10⁴ m/s
(d) M1 states energy conservation per unit mass: (1/2)*v_inf² + 0 = (1/2)*v_R² + (-GM/R)
(d) M1 correctly rearranges to v_R = √v_inf² + 2GM/R, equivalently v_R = √v_inf² + v_esc²
(d) A1 correctly substitutes v_inf = 1.60 x 10⁴ m/s and 2GM/R (or their v_esc from part (c))
(d) A1 v_R = 1.95 x 10⁴ m/s (awrt 3 sf), allow ecf
(d) Answer: v_R = 1.95 x 10⁴ m/s

Question 3

Level 1 (1-2): Makes only basic, largely isolated statements, for example simply asserting that a current is induced or that energy is 'lost' as heat, with little or no correct use of Faraday's law or Lenz's law and no clear chain of reasoning linking cause to effect.
Level 2 (3-4): Correctly identifies that the moving magnet causes a changing magnetic flux which induces eddy currents in the copper tube (Faraday), and that Lenz's law means these currents oppose the motion, producing a retarding force. The explanation of the energy transfer to heat is present but the reasoning chain (flux change, induced EMF, induced current, retarding force, reduced acceleration, energy dissipation) has gaps, is not fully sequenced, or contains minor errors. May not fully explain the plastic-tube case.
Level 3 (5-6): Gives a complete, correctly sequenced explanation: the falling magnet causes a changing flux linkage through the copper tube's wall, which by Faraday's law induces an EMF and hence eddy currents in the conducting tube; by Lenz's law these induced currents create a magnetic field that opposes the change producing them, so they exert an upward (retarding) force on the magnet, reducing its acceleration below g. Correctly explains that the magnet's lost gravitational potential energy is transferred, via the induced current, into heat (resistive/Joule heating) dissipated in the resistance of the tube. Correctly contrasts this with the plastic tube, where no closed conducting path exists, so no eddy currents (and hence no retarding force) are induced and the magnet falls with acceleration g. Uses correct scientific terminology throughout.
Indicative content:
- As the magnet falls, the magnetic flux linking any circular path within the wall of the copper tube changes with time.
- By Faraday's law, a changing flux linkage induces an EMF in the conducting tube.
- Because the copper tube forms continuous closed conducting loops, this induced EMF drives circulating (eddy) currents in the tube wall.
- By Lenz's law, the induced current acts in the direction that opposes the change that produced it, i.e. it opposes the relative motion between the magnet and the tube.
- This opposition means the eddy currents' magnetic field exerts a retarding (upward) force on the falling magnet.
- The net downward force on the magnet is therefore less than its weight, so its acceleration is less than g (and, given enough tube length, it approaches a constant terminal velocity).
- In the plastic tube there is no continuous conducting path, so no eddy currents (and therefore no retarding force) can be induced, and the magnet falls freely with acceleration g.
- By conservation of energy, the gravitational potential energy the magnet loses in the copper tube is transferred into electrical energy by the induced current, which is then dissipated as heat due to the resistance of the tube wall (rather than being converted entirely into kinetic energy, as in the plastic tube).

Question 4

(a) M1 correctly substitutes into F = Q1*Q2/(4*π*epsilon0*r²)
(a) A1 F = 2.40 x 10^-6 N (awrt 3 sf)
(a) B1 states the force is attractive, since P and Q carry opposite-sign charges
(a) Answer: F = 2.40 x 10^-6 N, attractive
(b) M1 correctly substitutes into E = Q(P)/(4*π*epsilon0*r²)
(b) A1 E = 333 N C^-1 (awrt 3 sf), allow 3.33 x 10² N/C
(b) Answer: E = 333 N C^-1
(c) M1 calculates the field due to P at the midpoint: E_P = 1.33 x 10³ N/C
(c) M1 calculates the field due to Q at the midpoint: E_Q = 2.00 x 10³ N/C
(c) A1 both individual values correct to 3 sf (allow ecf on method)
(c) A1 adds the two fields (since they act in the same direction) to give E_total = 3.33 x 10³ N/C (awrt)
(c) Answer: E_total = 3.33 x 10³ N C^-1
(d) B1 states the field points from P towards Q
(d) Answer: From P towards Q

Question 5

(a) B1 states the equilibrium condition q*E = m*g
(a) M1 rearranges to E = m*g/q and substitutes the given values
(a) A1 E = 3.00 x 10⁵ N/C (cao), matching the given value
(a) Answer: E = 3.00 x 10⁵ N C^-1
(b) M1 uses V = E*d
(b) A1 V = 1.50 x 10³ V (awrt, accept 1500 V), allow ecf
(b) Answer: V = 1.50 x 10³ V
(c) B1 recognises that, for a negative charge, the electric force is opposite to the field direction, so the field must point downward for the force on the drop to act upward
(c) B1 states the upper plate must be at the higher potential, since field lines point from high to low potential
(c) Answer: The upper plate is at the higher potential.

Question 6

(a) B1 C23 = C2 + C3 = 450 uF
(a) Answer: C23 = 450 uF
(b) M1 uses 1/C(total) = 1/C1 + 1/C23
(b) A1 correctly substitutes C1 = 150 uF and their C23 = 450 uF
(b) A1 C(total) = 112.5 uF (awrt), allow ecf
(b) Answer: C(total) = 112.5 uF
(c) M1 uses Q = C(total) x V
(c) A1 Q = 1.69 x 10^-3 C (awrt, allow 1.69 mC), allow ecf
(c) Answer: Q = 1.69 x 10^-3 C
(d) M1 uses V1 = Q/C1, using the total charge from part (c) (charge is the same through both series elements)
(d) A1 V1 = 11.3 V (awrt), allow ecf
(d) Answer: V1 = 11.3 V
(e) M1 uses W = (1/2)*C(total)*V²
(e) A1 correctly substitutes C(total) = 112.5 uF and V = 15.0 V
(e) A1 W = 1.27 x 10^-2 J (awrt, allow 12.7 mJ), allow ecf
(e) Answer: W = 1.27 x 10^-2 J

Question 7

(a) M1 uses time constant = R*C
(a) A1 tau = 14.1 s (awrt), correct unit s
(a) Answer: tau = 14.1 s
(b) M1 uses Q0 = C*V0
(b) A1 Q0 = 2.40 x 10^-3 C (awrt, allow 2.40 mC)
(b) Answer: Q0 = 2.40 x 10^-3 C
(c) M1 uses V = V0*exp(-t/(RC)) with their tau from part (a)
(c) A1 correctly substitutes t = 20.0 s and tau = 14.1 s
(c) A1 V = 1.94 V (awrt, allow ecf)
(c) Answer: V = 1.94 V
(d) B1 plots ln(V) on the y-axis against t on the x-axis
(d) B1 recognises that ln(V) = ln(V0) - t/(RC), so a straight line is obtained (of gradient -1/(RC))
(d) B1 states that RC is found from RC = -1/gradient (i.e. the negative reciprocal of the gradient of the line of best fit)
(d) Answer: Plot ln(V) against t; the gradient of the resulting straight line equals -1/(RC), so RC = -1/gradient.
(e) B1 e.g. start the timer/data logger at the exact instant the discharge begins (using a switch synchronised with the logger), or ensure the capacitor is fully charged to V0 before starting the discharge
(e) Answer: Synchronise the start of timing precisely with the instant the switch is used to begin the discharge (or ensure the capacitor is fully charged to V0 first).

Question 8

(a) M1 finds the change in mass reading: delta m = 152.87 - 152.40 = 0.47 g = 4.70 x 10^-4 kg
(a) M1 uses F = delta m * g
(a) A1 F = 4.61 x 10^-3 N (awrt 3 sf)
(a) Answer: F = 4.61 x 10^-3 N
(b) M1 uses B = F/(I*L)
(b) A1 B = 1.46 x 10^-2 T (awrt, allow 14.6 mT), allow ecf
(b) Answer: B = 1.46 x 10^-2 T
(c) B1 states the graph should be a straight line through the origin, with gradient equal to B*L
(c) B1 states B is found from B = gradient/L
(c) B1 explains that using the gradient of a line of best fit through several data points averages out random errors, giving a more reliable value than relying on a single (potentially anomalous) measurement
(c) Answer: Plot F against I; the gradient equals B*L, so B = gradient/L. Using a line of best fit through many points reduces the effect of random error compared with a single reading.

Question 9

(a) M1 uses q*V = (1/2)*m*v²
(a) A1 correctly rearranges to v = √2*q*V/m
(a) A1 v = 1.79 x 10⁵ m/s (awrt 3 sf)
(a) Answer: v = 1.79 x 10⁵ m/s
(b) M1 uses r = m*v/(B*q)
(b) A1 correctly substitutes m, their v from part (a), B and q
(b) A1 r = 4.96 x 10^-2 m (awrt), allow ecf
(b) Answer: r = 4.96 x 10^-2 m
(c) M1 recognises from q*V = (1/2)*m*v² that v is proportional to 1/√m, so the more massive carbon-13 ion has a smaller speed
(c) M1 recognises from r = m*v/(B*q) that, since v is proportional to 1/√m, r is proportional to m/√m = √m
(c) A1 concludes the carbon-13 ion has a greater mass, so √m is greater, so its radius is greater than that of the carbon-12 ion
(c) Answer: The carbon-13 ion has a greater radius than the carbon-12 ion.

Question 10

(a) M1 substitutes into EMF0 = B*A*N*ω
(a) A1 EMF0 = 74.3 V (awrt 3 sf)
(a) Answer: EMF0 = 74.3 V
(b) M1 uses f = ω/(2*π)
(b) A1 f = 47.7 Hz (awrt 3 sf)
(b) Answer: f = 47.7 Hz
(c) B1 states the EMF varies sinusoidally with time, reaching peak values of about +74.3 V and -74.3 V, with period T = 2*π/ω = 0.0209 s
(c) B1 states the EMF is zero twice per rotation, at the instants when the flux linkage through the coil is at a maximum or a minimum (a turning point of the flux linkage)
(c) B1 explains that at these instants the rate of change of flux linkage is (momentarily) zero, and since the induced EMF is proportional to the rate of change of flux linkage, the EMF is also zero at these instants
(c) Answer: The EMF varies sinusoidally with time (period 0.0209 s, peak ±74.3 V) and is zero twice per rotation, when the flux linkage is at a maximum or minimum, because the rate of change of flux linkage is momentarily zero at those instants.

Question 11

(a) M1 uses Ns = Np*(Vs/Vp)
(a) A1 Ns = 240 turns
(a) Answer: Ns = 240 turns
(b) M1 calculates output power: P(out) = Vs*Is = 96.0 W
(b) M1 uses efficiency = P(out)/P(in) to find P(in) = P(out)/0.920
(b) A1 P(in) = 104 W (awrt)
(b) A1 Ip = P(in)/Vp = 0.454 A (awrt), allow ecf
(b) Answer: Ip = 0.454 A

Question 12

(a) M1 equates the electric force q*E to the magnetic force q*v*B for an undeviated ion
(a) A1 correctly cancels q and rearranges to v = E/B
(a) A1 v = 5.00 x 10⁵ m/s
(a) Answer: v = 5.00 x 10⁵ m/s
(b) M1 uses r = m*v/(q*B), rearranged to q/m = v/(r*B)
(b) A1 correctly substitutes v = 5.00 x 10⁵ m/s (from part (a)), r = 4.20 x 10^-2 m and B = 0.350 T
(b) A1 q/m = 3.40 x 10⁷ C/kg (awrt), allow ecf
(b) Answer: q/m = 3.40 x 10⁷ C kg^-1
(c) B1 identifies that the field lines spread out (fringing/edge effects) near the edges of the parallel plates, so the field is not perfectly uniform close to the plate edges
(c) B1 states a suitable consequence, e.g. ions travelling near the edges experience a slightly different force balance and are not perfectly velocity-selected, so the emerging beam has a small spread of speeds rather than being purely monochromatic
(c) Answer: Fringing (edge) effects mean the field is weaker or less uniform near the plate edges; ions passing close to the edges are not perfectly velocity-selected, so the emerging beam has a small spread of speeds.

Physics: Fields (Gravitational, Electric, Magnetic) - Worksheets, Questions and Revision