D: Fields, motion in EM fields & induction – IB Physics exam questions with mark scheme

1Core3 marks

Mars has mass 6.42 × 10²³ kg and radius 3.39 × 10⁶ m.

(a) Define gravitational field strength at a point. [1]

(b) Calculate the gravitational field strength at the surface of Mars. [2]

(a) Force per unit mass acting on a (small) test mass placed at that point(1)
(b) Uses g = GM/R² with correct substitution (6.67 × 10⁻¹¹ × 6.42 × 10²³ ÷ (3.39 × 10⁶)²)(1)
(b) g = 3.73 N kg⁻¹ (allow m s⁻²)(1)
Answer: (a) force per unit mass on a test mass; (b) 3.73 N kg⁻¹
Examiner tip: Define it as force per unit mass – 'force on an object' alone does not earn the mark.

2Core4 marks

Two parallel metal plates are 12 mm apart in a vacuum and connected to a 240 V supply, as shown.

(a) Calculate the magnitude of the electric field strength between the plates. [2]

(b) An electron is released at rest midway between the plates. Calculate the force on it and state its direction. [2]

(a) E = V/d with d converted to metres (0.012 m)(1)
(a) E = 2.0 × 10⁴ V m⁻¹ (N C⁻¹)(1)
(b) F = eE = 1.60 × 10⁻¹⁹ × 2.0 × 10⁴ = 3.2 × 10⁻¹⁵ N(1)
(b) Direction: towards the positive (upper) plate, i.e. opposite to the field(1)
Answer: (a) 2.0 × 10⁴ V m⁻¹; (b) 3.2 × 10⁻¹⁵ N towards the positive plate
Examiner tip: Convert mm to m first – a factor of 1000 is the usual lost mark; an electron moves against the field.

3Standard4 marks

A satellite orbits the Earth in a circular orbit at a height of 600 km above the surface.

Mass of Earth = 5.97 × 10²⁴ kg; radius of Earth = 6.37 × 10⁶ m.

(a) Show that the orbital speed is about 7.6 km s⁻¹. [3]

(b) Calculate the orbital period in seconds. [1]

(a) Gravitational force provides the centripetal force: GMm/r² = mv²/r(1)
(a) Orbital radius r = 6.37 × 10⁶ + 6.0 × 10⁵ = 6.97 × 10⁶ m, and v = √(GM/r)(1)
(a) v = 7.56 × 10³ m s⁻¹ ≈ 7.6 km s⁻¹ (must show an unrounded value)(1)
(b) T = 2πr/v = 5.79 × 10³ s (5794 s; allow 5.8 × 10³ s)(1)
Answer: v = 7.56 × 10³ m s⁻¹; T = 5.79 × 10³ s
Examiner tip: Orbital radius is measured from Earth's centre: add the height to the radius of the Earth.

4StandardHL only5 marks

The Moon has mass 7.35 × 10²² kg and radius 1.74 × 10⁶ m.

(a) Calculate the gravitational potential at the Moon's surface. [2]

(b) Calculate the escape speed from the Moon's surface. [2]

(a) V = −GM/R with correct substitution(1)
(a) V = −2.82 × 10⁶ J kg⁻¹ (negative sign required)(1)
(b) Kinetic energy at surface equals the energy needed to reach V = 0: ½mv² = GMm/R (so v = √(2GM/R))(1)
(b) v = 2.37 × 10³ m s⁻¹(1)
(c) Mass m cancels from ½mv² = GMm/R (both KE and the change in potential energy are proportional to m)(1)
Answer: (a) −2.82 × 10⁶ J kg⁻¹; (b) 2.37 × 10³ m s⁻¹; (c) m cancels
Examiner tip: Gravitational potential is always negative (zero at infinity) – a missing minus sign loses the mark.

5Standard5 marks

A proton enters a region of uniform magnetic field of flux density 0.40 T at a speed of 2.5 × 10⁶ m s⁻¹. The velocity is perpendicular to the field, which is directed into the page.

(a) Calculate the magnetic force on the proton. [2]

(b) Calculate the radius of the proton's path. [2]

(a) F = qvB with correct substitution(1)
(a) F = 1.6 × 10⁻¹³ N(1)
(b) Magnetic force provides centripetal force: qvB = mv²/r so r = mv/(qB)(1)
(b) r = 0.065 m (6.5 × 10⁻² m)(1)
(c) Force is always perpendicular to velocity, so no work is done / speed unchanged(1)
Answer: (a) 1.6 × 10⁻¹³ N; (b) 0.065 m; (c) force ⟂ velocity so no work done
Examiner tip: Say the force is perpendicular to the motion so does no work – 'speed is constant' alone just restates the question.

6HardHL only5 marks

A coil of 100 turns has a magnetic flux Φ passing through each turn that varies with time as shown.

(a) Calculate the magnitude of the induced emf between 0 and 0.20 s. [2]

(b) State the induced emf between 0.20 s and 0.40 s and give a reason. [1]

(c) Calculate the magnitude of the induced emf between 0.40 s and 0.50 s and state how its polarity compares with that in (a). [2]

(a) ε = N ΔΦ/Δt (Faraday's law), using gradient 4.0 × 10⁻³ Wb ÷ 0.20 s(1)
(a) ε = 100 × 0.020 = 2.0 V(1)
(b) Zero, because the flux is constant so there is no change in flux linkage(1)
(c) ε = 100 × 4.0 × 10⁻³ ÷ 0.10 = 4.0 V(1)
(c) Polarity is opposite (flux is decreasing rather than increasing, so Lenz's law gives a reversed current/emf)(1)
Answer: (a) 2.0 V; (b) 0 V; (c) 4.0 V, opposite polarity
Examiner tip: Convert mWb to Wb and remember the number of turns – emf depends on flux linkage NΦ.

7HardHL only5 marks

A rectangular coil of 50 turns and area 0.020 m² rotates at a constant frequency of 50 Hz in a uniform magnetic field of flux density 0.15 T, forming a simple AC generator. The axis of rotation is perpendicular to the field.

(a) Calculate the maximum magnetic flux linkage through the coil. [1]

(b) The coil turns through a quarter of a revolution, starting from the position where its plane is perpendicular to the field. Calculate the average emf induced during this quarter turn. [2]

(d) The frequency of rotation is doubled. State the effect on the peak emf and on the period of the output. [1]

(a) NΦ = NBA = 50 × 0.15 × 0.020 = 0.15 Wb (accept Wb-turns)(1)
(b) Time for a quarter turn = T/4 = (1/50)/4 = 5.0 × 10⁻³ s and ΔΦ goes from BA to 0 (Δ(NΦ) = 0.15 Wb)(1)
(b) ε = NΔΦ/Δt = 0.15 ÷ 5.0 × 10⁻³ = 30 V (allow ECF from (a))(1)
(c) The plane of the coil is parallel to the field (accept: normal to the coil perpendicular to B; flux linkage zero but changing at its fastest rate)(1)
(d) Peak emf doubles and the period halves (both needed)(1)
Answer: (a) 0.15 Wb; (b) 30 V; (c) coil plane parallel to B; (d) peak emf ×2, period ÷2
Examiner tip: A quarter turn takes T/4 = 1/(4f); the emf is greatest when the flux linkage is zero, because that is when it is changing fastest.

8Hard5 marks

In a mass spectrometer, singly charged argon-40 ions (mass 6.64 × 10⁻²⁶ kg) are accelerated from rest through a potential difference of 2.0 kV and then enter a region of uniform magnetic field of flux density 0.25 T, perpendicular to their velocity.

(a) Calculate the speed of the ions on entering the field. [2]

(b) Calculate the radius of their circular path. [2]

(a) Gain in kinetic energy = work done by field: qV = ½mv²(1)
(a) v = 9.8 × 10⁴ m s⁻¹ (9.82 × 10⁴)(1)
(b) qvB = mv²/r so r = mv/(qB)(1)
(b) r = 0.16 m (allow 0.163 m)(1)
(c) r = 0.15 m (allow 0.155 m; allow 0.16 × √(5.98/6.64) = 0.152 m by ECF); r ∝ √m, so smaller than for Ar-40(1)
Answer: (a) 9.8 × 10⁴ m s⁻¹; (b) 0.16 m; (c) 0.15 m
Examiner tip: Use qV = ½mv² (not mv²); for the isotope in (c), r ∝ √m for the same charge, p.d. and B, so the lighter ion has the smaller radius.

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