Fields and their Consequences - Worksheets, Questions and Revision

12 original exam-style questions - 6 pages of questions with a full mark scheme - free printable PDF.

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A-Level · Physics

AP7 Fields and their Consequences

AQA 7407/7408 · Calculator allowed · about 150 minutes

Total Marks

Name: _______________________________ Date: ____ / ____ / ______

Answer ALL questions. Show all your working.

The dwarf planet-sized moon Phobos aside, Mars has mass 6.42 x 10²³ kg and radius 3.39 x 10⁶ m. Use the Physics Equations Sheet where needed; any equations not on the sheet are given below. Take G = 6.67 x 10^-11 N m² kg^-2.

(a)State Newton's law of gravitation for two point masses.(1)

(b)Show that the gravitational field strength at the surface of Mars is about 3.7 N/kg. Use g = GM/r².(3)

(c)A small rock is dropped from rest 2.0 m above the Martian surface. Assuming g = 3.7 N/kg is constant over this height, calculate the time taken for the rock to reach the surface. Use s = (1/2)g t².(3)

(d)Describe the shape of the gravitational field around Mars as a whole, and state one way this differs from the field close to a small region of the Martian surface (such as the 2.0 m drop in part (c)).(2)

(Total for Question 1 is 9 marks)

Continue to use the Mars data from Question 1 (mass 6.42 x 10²³ kg, radius 3.39 x 10⁶ m, G = 6.67 x 10^-11 N m² kg^-2).

(a)State what is meant by the gravitational potential at a point in a gravitational field.(1)

(b)Calculate the gravitational potential at the surface of Mars. Use V = -GM/r.(2)

(c)A spacecraft of mass 850 kg leaves the surface of Mars and just escapes its gravitational field. Using v = √2GM/r for the minimum escape velocity, calculate this velocity.(3)

(d)Explain, in terms of energy, why v = √2GM/r is called the escape velocity.(2)

(Total for Question 2 is 8 marks)

A Mars communications satellite moves in a circular orbit of radius r about the centre of Mars, held in orbit by the gravitational force. Continue to use GM(Mars) = 4.28 x 10¹³ N m² kg^-1.

(a)By equating the gravitational force GMm/r² to the centripetal force m(4 π² r)/T², show that the orbital period T of the satellite is given by T² = (4 π² r³) / (GM).(3)

(b)The satellite orbits at a radius of 2.0 x 10⁷ m from the centre of Mars. Calculate its orbital period, giving your answer in hours.(4)

(c)Mars rotates on its axis once every 24.6 hours. State what is meant by a synchronous orbit, and use your answer to part (b) to decide, with a reason, whether this satellite is in a synchronous orbit around Mars.(2)

(Total for Question 3 is 9 marks)

Two isolated point charges are fixed 0.15 m apart in air: charge A = +3.2 x 10^-9 C and charge B = -5.0 x 10^-9 C. Take k = 1/(4 π epsilon0) = 8.99 x 10⁹ N m² C^-2.

(a)Calculate the magnitude of the electrostatic force between charges A and B, stating whether the force is attractive or repulsive. Use F = Q1 Q2 / (4 π epsilon0 r²).(3)

(b)Calculate the electric field strength due to charge A alone at the position of charge B. Use E = kQ/r².(2)

(c)Describe the pattern of the electric field around an isolated positive point charge, including the direction of the field.(2)

(d)A proton (charge +1.6 x 10^-19 C) is placed at the midpoint between A and B, 0.075 m from each charge. Calculate the resultant electrostatic force on the proton, explaining your reasoning about direction.(4)

(Total for Question 4 is 11 marks)

Continue with charge A = +3.2 x 10^-9 C from Question 4.

(a)State what is meant by the electric potential at a point in an electric field.(1)

(b)Calculate the electric potential at a point 0.075 m from charge A. Use V = kQ/r.(2)

(c)A proton (charge +1.6 x 10^-19 C) is brought from a very large distance to this point. Calculate the work done on the proton. Use W = qV.(2)

(d)Compare and contrast gravitational fields and electric fields. Your answer should refer to the mathematical form of the force, field strength and potential, and to the nature of the sources that produce each field.(6)

(Total for Question 5 is 11 marks)

Two parallel plates are separated by 0.020 m, with a potential difference of 500 V between them, creating a uniform electric field. The plates have a length of 0.040 m (measured along the direction of travel of a particle passing between them). An electron (mass 9.11 x 10^-31 kg, charge 1.60 x 10^-19 C) enters midway between the plates, travelling parallel to them at 2.0 x 10⁷ m/s.

(a)Calculate the electric field strength between the plates. Use E = V/d.(2)

(b)(i) Calculate the electric force on the electron while it is between the plates. Use F = eE.(2)

(c)(ii) Calculate the vertical acceleration of the electron while it is between the plates. Use a = F/m.(2)

(d)(iii) Calculate the time the electron spends between the plates. Use t = L/v, where L is the plate length.(2)

(e)(iv) Calculate the vertical deflection of the electron as it leaves the plates, using y = (1/2)a t². Hence determine, with a reason, whether the electron leaves the plates without striking either one (the maximum possible deflection before hitting a plate is half the plate separation).(3)

(Total for Question 6 is 11 marks)

A capacitor of capacitance 470 uF is charged until the potential difference across it is 9.0 V.

(a)State what is meant by the capacitance of a capacitor.(1)

(b)Calculate the charge stored on the capacitor. Use Q = CV.(2)

(c)Calculate the energy stored in the capacitor. Use W = (1/2)CV².(2)

(d)This 470 uF capacitor is now connected in series with a 220 uF capacitor, and the combination is charged by a 9.0 V supply. Calculate the total capacitance of the combination and hence the charge stored.(4)

(Total for Question 7 is 9 marks)

REQUIRED PRACTICAL. A student investigates the discharge of a capacitor of capacitance C through a resistor of resistance R = 15 kOhm. The capacitor is charged to V0 = 6.0 V and then allowed to discharge, with the pd V across the capacitor recorded at intervals using a voltmeter and a stopwatch. The equation for the discharge is V = V0 exp(-t/RC). Results are shown below.
t / s: 0, 20, 40, 60, 80, 100
V / V: 6.0, 3.3, 1.8, 0.97, 0.53, 0.29

(a)Explain how the equation V = V0 exp(-t/RC) can be rearranged into a form suitable for plotting a straight-line graph, stating what should be plotted on each axis and what the gradient represents.(3)

(b)Using the table of results, calculate ln(V) for t = 0 s and for t = 100 s, giving each answer to 2 decimal places.(2)

(c)Using your two values of ln(V) from part (b), calculate the gradient of the ln(V) against t graph, and hence determine the time constant RC of the circuit.(4)

(d)The resistor used has resistance 15 kOhm. Calculate the capacitance of the capacitor, giving your answer to an appropriate number of significant figures.(2)

(e)State one source of systematic error in this experiment and suggest a way of reducing it.(2)

(Total for Question 8 is 13 marks)

A straight wire of length 0.25 m carries a current of 4.5 A. It is placed perpendicular to a uniform magnetic field of flux density 0.18 T.

(a)Calculate the force on the wire due to the magnetic field. Use F = BIL.(2)

(b)The current flows from left to right and the magnetic field points into the page. State the direction of the force on the wire and explain how you determined it.(2)

(c)The wire is now turned so that it makes an angle of 30 degrees with the magnetic field (rather than being perpendicular to it). Calculate the new force on the wire. Use F = BIL sin(θ).(3)

(Total for Question 9 is 7 marks)

REQUIRED PRACTICAL. A student uses a current balance to determine the magnetic flux density between two magnadur magnets. A stiff horizontal wire of length 0.060 m passes between the magnets and rests on a top-pan balance support. When a current of 2.0 A is switched on, the balance reading changes by a mass equivalent of 0.24 g. Take g = 9.81 N/kg.

(a)Explain why switching on the current changes the reading on the balance.(2)

(b)Calculate the force on the wire corresponding to this change in balance reading. Use F = mg.(2)

(c)Calculate the magnetic flux density between the magnets. Use F = BIL.(3)

(d)The student repeats the measurement for several different currents and plots a graph of force F (y-axis) against current I (x-axis). Explain how the gradient of this graph is related to B, and why using the gradient gives a more reliable value of B than using a single pair of readings.(3)

(Total for Question 10 is 10 marks)

A flat coil of 250 turns has a cross-sectional area of 4.0 x 10^-3 m². Its plane is perpendicular to a uniform magnetic field. The flux density is increased steadily from 0.020 T to 0.080 T over a time of 0.50 s.

(a)Calculate the change in magnetic flux through the coil. Use flux = BA.(2)

(b)Calculate the average emf induced in the coil during this time. Use emf = N (delta flux) / (delta t).(3)

(c)State Lenz's law and use it to explain the direction in which the induced current acts in the coil while the flux is increasing.(3)

(d)The coil is connected to a resistor of resistance 15 Ohm. Calculate the average induced current during the 0.50 s interval. Use I = emf/R.(2)

(Total for Question 11 is 10 marks)

A proton (mass 1.67 x 10^-27 kg, charge 1.60 x 10^-19 C) travels at 2.5 x 10⁶ m/s perpendicular to a uniform magnetic field of flux density 0.40 T, which occupies a region of width 5.0 cm. The proton enters the field travelling perpendicular to the boundary of the field region.

(a)Calculate the magnetic force on the proton as it enters the field. Use F = BQv.(2)

(b)Explain why this magnetic force causes the proton to move at constant speed along a circular path.(2)

(c)By equating the magnetic force BQv to the centripetal force mv²/r, show that the radius of the circular path is given by r = mv/(BQ).(2)

(d)Calculate the radius of the circular path followed by this proton.(2)

(e)The magnetic field region has a width of 5.0 cm, measured in the direction the proton was initially travelling. For a particle entering perpendicular to the boundary of a field region, the maximum distance it can penetrate into the field before curving back out of the same boundary is equal to the radius r of its circular path. Using your value of r from part (d), determine, with reasoning, whether this proton crosses completely through the 5.0 cm field region or curves back out of the boundary it entered through.(3)

(Total for Question 12 is 11 marks)

Mark scheme · AP7 Fields and their Consequences

Question 1

(a) B1 the gravitational force between two point masses is directly proportional to the product of the masses and inversely proportional to the square of their separation (oe, e.g. F = GMm/r² with terms defined)
(a) Answer: F = GMm/r², an attractive force along the line joining the masses
(b) M1 correct substitution g = (6.67 x 10^-11 x 6.42 x 10²³) / (3.39 x 10⁶)²
(b) M1 numerator evaluated as awrt 4.28 x 10¹³ and denominator evaluated as awrt 1.15 x 10¹³ (oe combined arithmetic)
(b) A1 g = 3.7 N/kg (cso, to 2 sf, matches given value)
(b) Answer: g = 3.7 N/kg
(c) M1 rearrangement t = √2s/g
(c) M1 correct substitution t = √2 x 2.0 / 3.7
(c) A1 t = 1.0 s (awrt 1.0 s, cao)
(c) Answer: t = 1.04 s (awrt 1.0 s)
(d) B1 the field around the whole of Mars is radial, with field lines directed towards the centre of Mars, and field strength decreasing with distance from the centre
(d) B1 close to the surface over a small region the field is effectively uniform (parallel, equally spaced field lines) because the change in r is negligible (oe)
(d) Answer: Radial overall; uniform over a small region near the surface

Question 2

(a) B1 the work done per unit mass in bringing a small test mass from infinity to that point (oe)
(a) Answer: Work done per unit mass moving a small mass from infinity to the point
(b) M1 correct substitution V = -(6.67 x 10^-11 x 6.42 x 10²³) / (3.39 x 10⁶)
(b) A1 V = -1.3 x 10⁷ J/kg (awrt -1.3 x 10⁷ J/kg, cao)
(b) Answer: V = -1.26 x 10⁷ J/kg (awrt -1.3 x 10⁷ J/kg)
(c) M1 correct substitution v = √2 x 6.67 x 10^-11 x 6.42 x 10²³ / 3.39 x 10⁶
(c) M1 correct evaluation inside the square root, awrt 2.5 x 10⁷
(c) A1 v = 5.0 x 10³ m/s (awrt 5.0 x 10³ m/s, cao)
(c) Answer: v = 5.0 x 10³ m/s (5.0 km/s)
(d) B1 at this speed the kinetic energy of the object equals the magnitude of the gravitational potential energy at that point (oe)
(d) B1 so the object has just enough energy to reach an infinite distance (where gravitational PE = 0) with zero final kinetic energy, i.e. it is not pulled back (oe)
(d) Answer: KE at launch equals magnitude of GPE, so the object can just reach infinity with zero remaining speed

Question 3

(a) M1 correctly equates GMm/r² = m(4 π² r)/T²
(a) M1 cancels m from both sides and rearranges to isolate T² (e.g. GM/r² = 4 π² r / T²)
(a) A1 T² = 4 π² r³ / (GM) (cso, matches the given expression)
(a) Answer: T² = 4 π² r³ / (GM)
(b) M1 correct substitution T² = 4 π² x (2.0 x 10⁷)³ / (4.28 x 10¹³)
(b) M1 T² evaluated as awrt 7.4 x 10⁹ s²
(b) M1 T = √T² evaluated as awrt 8.6 x 10⁴ s
(b) A1 T = 23.9 hours (awrt 23.9 hours, cao, after dividing by 3600)
(b) Answer: T = 8.59 x 10⁴ s = 23.9 hours
(c) B1 a synchronous orbit is one where the orbital period equals the rotation period of the planet, so the satellite remains above the same point (oe)
(c) B1 not synchronous, since 23.9 hours does not equal 24.6 hours; a slightly larger orbital radius would be needed to increase T to 24.6 hours (oe, allow ft on T from (b))
(c) Answer: Not synchronous: T = 23.9 h is close to but not equal to Mars's 24.6 h rotation period

Question 4

(a) M1 correct substitution F = 8.99 x 10⁹ x (3.2 x 10^-9 x 5.0 x 10^-9) / (0.15)²
(a) A1 F = 6.4 x 10^-6 N (awrt 6.4 x 10^-6 N, cao)
(a) B1 attractive, since the charges have opposite signs
(a) Answer: F = 6.4 x 10^-6 N, attractive
(b) M1 correct substitution E = 8.99 x 10⁹ x 3.2 x 10^-9 / (0.15)²
(b) A1 E = 1.3 x 10³ N/C (awrt 1.3 x 10³ N/C, cao)
(b) Answer: E = 1.28 x 10³ N/C (awrt 1.3 x 10³ N/C)
(c) B1 radial field, with field lines becoming more widely spaced further from the charge (field strength decreases with distance)
(c) B1 field lines point radially outwards, away from the positive charge
(c) Answer: Radial field pointing outwards from the charge, spacing increasing with distance
(d) M1 force due to A: F_A = 8.99 x 10⁹ x 3.2 x 10^-9 x 1.6 x 10^-19 / (0.075)² = 8.2 x 10^-16 N (awrt), directed away from A (repulsive, A positive)
(d) M1 force due to B: F_B = 8.99 x 10⁹ x 5.0 x 10^-9 x 1.6 x 10^-19 / (0.075)² = 1.3 x 10^-15 N (awrt), directed towards B (attractive, B negative)
(d) M1 recognises both forces act in the same direction (from A towards B) at the midpoint, so they add
(d) A1 resultant force = 2.1 x 10^-15 N (awrt 2.1 x 10^-15 N), directed from A towards B
(d) Answer: F_resultant = 2.1 x 10^-15 N, directed from A towards B

Question 5

(a) B1 the work done per unit positive charge in bringing a small test charge from infinity to that point (oe)
(a) Answer: Work done per unit positive charge moving a small charge from infinity to the point
(b) M1 correct substitution V = 8.99 x 10⁹ x 3.2 x 10^-9 / 0.075
(b) A1 V = 3.8 x 10² V (awrt 380 V, cao)
(b) Answer: V = 384 V (awrt 380 V)
(c) M1 correct substitution W = 1.6 x 10^-19 x 383.6 (ft from (b))
(c) A1 W = 6.1 x 10^-17 J (awrt 6.1 x 10^-17 J, ft from (b))
(c) Answer: W = 6.14 x 10^-17 J (awrt 6.1 x 10^-17 J)
(d) L1 (1-2): Basic, isolated statements about gravitational or electric fields with little or no valid comparison. Limited use of correct physics terminology.
(d) L2 (3-4): Some valid points of comparison and/or contrast are made, with partially correct links between force, field strength and potential for both field types. Explanation lacks full coherence.
(d) L3 (5-6): A clear, well-linked comparison covering the inverse-square form of both forces, the definitions of field strength and potential for each field, and the key difference arising from the sign of the sources (mass vs charge). The answer is coherent and uses correct physics terminology throughout.
(d) Indicative content:
- Both forces obey an inverse square law: F = GMm/r^2 for gravity and F = Qq/(4 pi epsilon0 r^2) for electric charges.
- Gravitational field strength g = GM/r^2 and electric field strength E = kQ/r^2 have the same mathematical form, force per unit mass and force per unit charge respectively.
- Gravitational potential V = -GM/r is always negative (taking zero at infinity) because mass is always positive and the force is always attractive.
- Electric potential V = kQ/r can be positive or negative depending on the sign of the source charge, because charge can be positive or negative.
- Gravitational force is always attractive because there is no negative mass; electric force can be attractive or repulsive depending on the signs of the two charges.
- Both fields can be represented by radial field lines around a point source, and both have field strength related to the gradient of the potential.
- Gravitational fields are generally far weaker than electric fields for comparable point sources, because G is very much smaller in magnitude than k = 1/(4 pi epsilon0).
(d) Answer: Levels-of-response: see indicative content

Question 6

(a) M1 correct substitution E = 500/0.020
(a) A1 E = 2.5 x 10⁴ V/m (cao)
(a) Answer: E = 2.5 x 10⁴ V/m
(b) M1 correct substitution F = 1.6 x 10^-19 x 2.5 x 10⁴ (ft from (a))
(b) A1 F = 4.0 x 10^-15 N (cao, ft from (a))
(b) Answer: F = 4.0 x 10^-15 N
(c) M1 correct substitution a = 4.0 x 10^-15 / 9.11 x 10^-31 (ft from (b))
(c) A1 a = 4.4 x 10¹⁵ m/s² (awrt 4.4 x 10¹⁵ m/s², ft from (b))
(c) Answer: a = 4.39 x 10¹⁵ m/s² (awrt 4.4 x 10¹⁵ m/s²)
(d) M1 correct substitution t = 0.040 / 2.0 x 10⁷
(d) A1 t = 2.0 x 10^-9 s (cao)
(d) Answer: t = 2.0 x 10^-9 s
(e) M1 correct substitution y = 0.5 x 4.4 x 10¹⁵ x (2.0 x 10^-9)² (ft from (c) and (d))
(e) A1 y = 8.8 x 10^-3 m (awrt 8.8 mm, ft)
(e) B1 compares y to the maximum allowed deflection of 0.010 m (half the 0.020 m gap) and correctly concludes that the electron leaves the plates without striking one, since 8.8 mm < 10 mm
(e) Answer: y = 8.8 x 10^-3 m; electron exits without hitting a plate (8.8 mm < 10 mm maximum)

Question 7

(a) B1 the charge stored per unit potential difference across the capacitor (oe, C = Q/V)
(a) Answer: Charge stored per unit potential difference, C = Q/V
(b) M1 correct substitution Q = 470 x 10^-6 x 9.0
(b) A1 Q = 4.2 x 10^-3 C (awrt 4.2 x 10^-3 C, cao)
(b) Answer: Q = 4.23 x 10^-3 C (awrt 4.2 mC)
(c) M1 correct substitution W = 0.5 x 470 x 10^-6 x 9.0²
(c) A1 W = 1.9 x 10^-2 J (awrt 1.9 x 10^-2 J, cao)
(c) Answer: W = 1.90 x 10^-2 J (19 mJ)
(d) M1 correct use of series formula 1/C = 1/470 + 1/220 (in uF, or SI equivalent)
(d) A1 C = 1.5 x 10^-4 F (awrt 150 uF, cao)
(d) M1 correct substitution Q = C x V using their C (ft)
(d) A1 Q = 1.3 x 10^-3 C (awrt 1.3 x 10^-3 C, ft)
(d) Answer: C = 150 uF (1.5 x 10^-4 F); Q = 1.35 x 10^-3 C

Question 8

(a) M1 taking natural logs of both sides gives ln V = ln V0 - t/(RC)
(a) A1 plot ln V (y-axis) against t (x-axis) to give a straight line
(a) A1 the gradient of the line is equal to -1/(RC), i.e. -1/tau
(a) Answer: ln V = ln V0 - t/(RC); plot ln V against t; gradient = -1/(RC)
(b) B1 ln(6.0) = 1.79
(b) B1 ln(0.29) = -1.24
(b) Answer: ln(6.0) = 1.79; ln(0.29) = -1.24
(c) M1 correct gradient calculation, gradient = (-1.24 - 1.79)/(100 - 0) (ft from (b))
(c) A1 gradient = -0.0303 s^-1 (awrt -0.030 s^-1, ft)
(c) M1 correct use of tau = -1/gradient
(c) A1 tau = 33 s (awrt 33 s, ft)
(c) Answer: gradient = -0.0303 s^-1; time constant RC = 33 s
(d) M1 correct rearrangement and substitution C = tau/R = 33/15000 (ft from (c))
(d) A1 C = 2.2 x 10^-3 F (2200 uF) (awrt 2.2 x 10^-3 F, 2 sf, ft)
(d) Answer: C = 2.2 x 10^-3 F (2200 uF)
(e) B1 human reaction time when starting/stopping the stopwatch or reading the voltmeter introduces a systematic timing error (oe)
(e) B1 use a data logger with a voltage sensor to record V automatically at precise time intervals, removing reliance on manual timing (oe)
(e) Answer: Reaction-time error in manual timing; use a data logger and voltage sensor instead

Question 9

(a) M1 correct substitution F = 0.18 x 4.5 x 0.25
(a) A1 F = 0.20 N (cao)
(a) Answer: F = 0.2025 N (awrt 0.20 N)
(b) B1 the force acts vertically upwards (in the plane of the page)
(b) B1 found using Fleming's left-hand rule: First finger = Field (into page), seCond finger = Current (left to right), thuMb = force/Motion, giving an upward force
(b) Answer: Force acts vertically upwards, found using Fleming's left-hand rule
(c) M1 correct substitution F = 0.18 x 4.5 x 0.25 x sin(30)
(c) M1 sin(30) = 0.5 used correctly
(c) A1 F = 0.10 N (cao)
(c) Answer: F = 0.101 N (awrt 0.10 N)

Question 10

(a) B1 the current-carrying wire in the magnetic field experiences a magnetic force (F = BIL)
(a) B1 by Newton's third law, an equal and opposite force acts on the magnets (and their support resting on the balance), changing the reading (oe)
(a) Answer: The wire experiences a magnetic force; the reaction force on the magnets changes the balance reading
(b) M1 correct substitution F = 0.24 x 10^-3 x 9.81
(b) A1 F = 2.4 x 10^-3 N (awrt 2.4 x 10^-3 N, cao)
(b) Answer: F = 2.35 x 10^-3 N (awrt 2.4 x 10^-3 N)
(c) M1 correct rearrangement B = F/(IL)
(c) M1 correct substitution B = 2.4 x 10^-3 / (2.0 x 0.060) (ft from (b))
(c) A1 B = 2.0 x 10^-2 T (awrt 0.020 T, ft)
(c) Answer: B = 1.96 x 10^-2 T (awrt 0.020 T)
(d) B1 since F = BIL, for constant L the gradient of F against I equals BL
(d) B1 so B = gradient / L
(d) B1 using a line of best fit through several points averages out random errors in individual readings, giving a more reliable value than a single measurement (oe)
(d) Answer: gradient = BL, so B = gradient/L; using several points reduces the effect of random error

Question 11

(a) M1 correct substitution using change in B: (0.080 - 0.020) x 4.0 x 10^-3
(a) A1 change in flux = 2.4 x 10^-4 Wb (cao)
(a) Answer: delta(flux) = 2.4 x 10^-4 Wb
(b) M1 correct substitution emf = 250 x 2.4 x 10^-4 / 0.50 (ft from (a))
(b) M1 flux linkage change N x (delta flux) evaluated as 0.060 Wb-turns
(b) A1 emf = 0.12 V (cao, ft)
(b) Answer: emf = 0.12 V
(c) B1 Lenz's law: the direction of an induced emf (and current) is such as to oppose the change that produces it (oe, consistent with conservation of energy)
(c) B1 since the flux through the coil is increasing, the induced current flows in a direction that creates its own magnetic field opposing this increase (i.e. a field in the opposite sense to the applied field, inside the coil)
(c) B1 the direction of current flow around the coil (e.g. clockwise or anticlockwise as viewed along the field) can then be found from the right-hand grip rule applied to the opposing field
(c) Answer: Induced current opposes the increasing flux; its direction found from the right-hand grip rule applied to the opposing field
(d) M1 correct substitution I = 0.12/15 (ft from (b))
(d) A1 I = 8.0 x 10^-3 A (cao, ft)
(d) Answer: I = 8.0 x 10^-3 A (8.0 mA)