Theme D · Fields

A planet has twice the mass and twice the radius of the Earth. The gravitational field strength at the Earth's surface is g. What is the field strength at the surface of the planet?

1. Ag/4
2. Bg/2
3. Cg
4. D2g

Two identical conducting spheres carry charges of +6.0 nC and −2.0 nC. They are touched together and then separated to their original distance apart. How does the magnitude of the force between them change?

1. AIt becomes zero.
2. BIt is reduced to one third of its original value.
3. CIt is reduced to one twelfth of its original value.
4. DIt increases to twelve times its original value.

A proton moves in a circular path of radius r in a uniform magnetic field of magnitude B, perpendicular to its velocity. If the speed of the proton is doubled, what is the new radius?

1. Ar/2
2. Br
3. C2r
4. D4r

A satellite of mass m is in a circular orbit of radius r around a planet of mass M. What is its total energy?

1. A−GMm/(2r)
2. B−GMm/r
3. C+GMm/(2r)
4. Dzero

A bar magnet is dropped, north pole downward, through a horizontal copper ring. Which describes the force on the magnet as it approaches the ring and after it has passed through?

1. Aupward as it approaches, upward after it has passed
2. Bupward as it approaches, downward after it has passed
3. Cdownward as it approaches, upward after it has passed
4. Dthere is no force, because copper is not magnetic

Which statement about equipotential surfaces and field lines is correct for any static field?

1. AThey are parallel to one another.
2. BThey are perpendicular to one another.
3. CWork is done moving a charge along a field line but not along an equipotential.
4. DBoth B and C are correct.

A student measures the magnetic flux density B at various perpendicular distances r from a long straight wire carrying a steady current I. Theory predicts B = μ₀I/(2πr).

1. (a)

   Explain Explain why a graph of B against 1/r is plotted rather than a graph of B against r.

   [2]
   Ready to self-mark?Reveal the detailed answerAO42 marks+

   Mark-by-mark answer

   1. 1

      B is inversely proportional to r, so a graph of B against r is a curve from which no constant can be read

   2. 2

      plotting against 1/r linearises the relationship, so the gradient gives μ₀I/(2π) directly and the straightness of the line tests the inverse proportionality

2. (b)

   Determine Determine the gradient of the line, and hence determine the current in the wire.

   [4]
   Ready to self-mark?Reveal the detailed answerAO24 marks+

   Mark-by-mark answer

   1. 1

      gradient = (50.0 − 10.0) × 10⁻⁶ / (50.0 − 10.0) = 1.0 × 10⁻⁶ T m

   2. 2

      A plot of B against 1/r has gradient = μ₀I/(2π).

   3. 3

      I = 2π × 1.0 × 10⁻⁶ / (4π × 10⁻⁷)

   4. 4

      I = 5.0 A

3. (c)

   Suggest The student's measurements at the largest distances scatter more about the line than those close to the wire. Suggest why.

   [2]
   Ready to self-mark?Reveal the detailed answerAO32 marks+

   Mark-by-mark answer

   1. 1

      the field is weakest at large r, so the reading is a smaller fraction of the probe's full scale and the percentage uncertainty in each reading is larger

   2. 2

      the Earth's magnetic field and any stray fields are a larger fraction of the measured value at those distances

4. (d)

   Outline Outline one systematic error that would make every value of B too large, and state how it would show up on the graph.

   [2]
   Ready to self-mark?Reveal the detailed answerAO32 marks+

   Mark-by-mark answer

   1. 1

      the probe not being zeroed before the measurements, or the Earth's field component along the probe not being subtracted, would add a constant to every reading

   2. 2

      the line would still be straight but would not pass through the origin — it would have a positive intercept on the B axis

A planet has mass 6.4 × 10²³ kg and radius 3.4 × 10⁶ m. The gravitational constant is 6.67 × 10⁻¹¹ N m² kg⁻².

1. (a)

   Calculate Calculate the gravitational field strength at the surface of the planet.

   [2]
   Ready to self-mark?Reveal the detailed answerAO22 marks+

   Mark-by-mark answer

   1. 1

      g = GM/R² = 6.67 × 10⁻¹¹ × 6.4 × 10²³ / (3.4 × 10⁶)²

   2. 2

      g = 3.7 N kg⁻¹

2. (b)

   Determine Determine the escape speed from the surface of the planet.

   [3]
   Ready to self-mark?Reveal the detailed answerAO23 marks+

   Mark-by-mark answer

   1. 1

      escape requires the total energy to be zero: ½mv² = GMm/R

   2. 2

      v = √(2GM/R) = √(2 × 6.67 × 10⁻¹¹ × 6.4 × 10²³ / 3.4 × 10⁶)

   3. 3

      v = 5.0 × 10³ m s⁻¹

3. (c)

   Explain Explain why gravitational potential is always negative, and state what a gravitational potential of −1.3 × 10⁷ J kg⁻¹ at the planet's surface means.

   [3]
   Ready to self-mark?Reveal the detailed answerAO23 marks+

   Mark-by-mark answer

   1. 1

      gravitational potential is defined as zero at infinity, and gravity is always attractive

   2. 2

      so bringing a mass in from infinity releases energy, and its potential energy must be less than zero everywhere else

   3. 3

      the value means that 1.3 × 10⁷ J of energy must be supplied to move each kilogram from the surface to infinity

4. (d)

   Discuss The planet has almost no atmosphere, while the Earth retains a substantial one. Discuss how the escape speed helps to explain this difference.

   [4]
   Ready to self-mark?Reveal the detailed answerAO34 marks+

   Mark-by-mark answer

   1. 1

      the molecules of a gas have a distribution of speeds, with some far above the mean at any temperature

   2. 2

      molecules whose speed exceeds the escape speed at the top of the atmosphere can leave the planet permanently

   3. 3

      this planet's escape speed of 5.0 km s⁻¹ is less than half the Earth's 11.2 km s⁻¹, so a much larger fraction of its molecules exceed it

   4. 4

      over geological time that fraction is lost, and the lightest gases go first — which is why any remaining atmosphere is thin and dominated by the heavier molecules

An electron is accelerated from rest through a potential difference of 500 V and then enters a region of uniform magnetic field of magnitude 2.5 mT, travelling perpendicular to the field. The mass of an electron is 9.11 × 10⁻³¹ kg and its charge has magnitude 1.60 × 10⁻¹⁹ C.

1. (a)

   Determine Determine the speed of the electron as it enters the magnetic field.

   [3]
   Ready to self-mark?Reveal the detailed answerAO23 marks+

   Mark-by-mark answer

   1. 1

      The gained electric potential energy becomes kinetic energy, so eV = ½mv².

   2. 2

      v = √(2 × 1.60 × 10⁻¹⁹ × 500 / 9.11 × 10⁻³¹)

   3. 3

      v = 1.33 × 10⁷ m s⁻¹

2. (b)

   Determine Determine the radius of the circular path followed by the electron in the field.

   [3]
   Ready to self-mark?Reveal the detailed answerAO23 marks+

   Mark-by-mark answer

   1. 1

      the magnetic force provides the centripetal force: evB = mv²/r

   2. 2

      r = mv/(eB) = 9.11 × 10⁻³¹ × 1.33 × 10⁷ / (1.60 × 10⁻¹⁹ × 2.5 × 10⁻³)

   3. 3

      r = 3.0 × 10⁻² m

3. (c)

   Explain Explain why the speed of the electron is unchanged while it moves within the magnetic field, even though its velocity is changing continuously.

   [3]
   Ready to self-mark?Reveal the detailed answerAO23 marks+

   Mark-by-mark answer

   1. 1

      the magnetic force is always perpendicular to the velocity

   2. 2

      so it has no component along the direction of motion and does no work on the electron

   3. 3

      the kinetic energy is therefore constant, and only the direction of the velocity changes

4. (d)

   Compare A uniform electric field is now applied in the same region instead of the magnetic field, perpendicular to the electron's initial velocity. Compare the resulting path and the resulting speed with those in the magnetic field.

   [3]
   Ready to self-mark?Reveal the detailed answerAO33 marks+

   Mark-by-mark answer

   1. 1

      in the electric field the force is constant in magnitude and direction, so the path is parabolic rather than circular

   2. 2

      the force has a component along the velocity once the electron has been deflected, so it does work and the speed increases

   3. 3

      in the magnetic field the force turns with the velocity, keeping the path circular and the speed constant

A flat rectangular coil of 250 turns and area 3.2 × 10⁻³ m² is rotated at a constant rate in a uniform magnetic field of magnitude 85 mT. The axis of rotation is perpendicular to the field. The coil completes 50 revolutions per second.

1. (a)

   State State Faraday's law of electromagnetic induction.

   [2]
   Ready to self-mark?Reveal the detailed answerAO12 marks+

   Mark-by-mark answer

   1. 1

      the magnitude of the induced e.m.f. is equal to the rate of change of magnetic flux linkage

   2. 2

      expressed as ε = −N dΦ/dt, the minus sign expressing Lenz's law

2. (b)

   Determine Determine the maximum e.m.f. induced in the coil.

   [3]
   Ready to self-mark?Reveal the detailed answerAO23 marks+

   Mark-by-mark answer

   1. 1

      ω = 2πf = 2π × 50 = 314 rad s⁻¹

   2. 2

      ε₀ = NBAω = 250 × 0.085 × 3.2 × 10⁻³ × 314

   3. 3

      ε₀ = 21 V

3. (c)

   Explain Explain why the induced e.m.f. is zero at the instant when the plane of the coil is perpendicular to the magnetic field, even though the flux through the coil is greatest at that moment.

   [3]
   Ready to self-mark?Reveal the detailed answerAO33 marks+

   Mark-by-mark answer

   1. 1

      the induced e.m.f. depends on the rate of change of flux, not on the flux itself

   2. 2

      when the plane of the coil is perpendicular to the field the flux is at a maximum, and at a maximum its rate of change is momentarily zero

   3. 3

      so the e.m.f. is zero there and greatest a quarter of a cycle later, when the flux is passing through zero and changing fastest

4. (d)

   Discuss The output of the coil is connected to a resistor. Discuss the energy transfers taking place, and explain why more mechanical power must be supplied to keep the coil rotating at 50 revolutions per second once the resistor is connected.

   [4]
   Ready to self-mark?Reveal the detailed answerAO34 marks+

   Mark-by-mark answer

   1. 1

      with the circuit closed, the induced e.m.f. drives a current through the coil and the resistor, and electrical energy is dissipated in the resistor as thermal energy

   2. 2

      the current-carrying coil sits in the magnetic field, so it experiences a torque

   3. 3

      by Lenz's law that torque opposes the rotation of the coil

   4. 4

      so an external agent must do work against it, and the mechanical power supplied equals the electrical power dissipated — energy is conserved, with the coil acting as the converter rather than the source