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A-Level Physics – Fields

Gravitational, electric and magnetic fields, induction.

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12 questions  ·  49 marks  ·  answers included  ·  A-Level (Edexcel/AQA style)
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This worksheet covers:

  • This worksheet covers gravitational
  • Electric and magnetic fields
  • Electromagnetic induction

A calculator is required; use $g = 9.81\,\text{m s}^{-2}$, $G = 6.67\times10^{-11}\,\text{N m}^2\text{kg}^{-2}$, $k = 8.99\times10^{9}\,\text{N m}^2\text{C}^{-2}$ and $e = 1.60\times10^{-19}\,\text{C}$ unless stated otherwise. Allow about 45 minutes.

Questions

Section A – Gravitational fields

1

A planet has mass $6.42\times10^{23}\,\text{kg}$ and radius $3.39\times10^{6}\,\text{m}$.(a) State the equation for gravitational field strength at the surface of a planet.(b) Calculate the gravitational field strength at the planet’s surface.

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(a) $g = \dfrac{GM}{r^2}$(b) $g = \dfrac{(6.67\times10^{-11})(6.42\times10^{23})}{(3.39\times10^{6})^2} = 3.73\,\text{m s}^{-2}$ (3 s.f.)
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2

A satellite orbits Earth (mass $5.97\times10^{24}\,\text{kg}$) in a circular orbit of radius $4.22\times10^{7}\,\text{m}$.(a) Show that the orbital speed is given by $v = \sqrt{\dfrac{GM}{r}}$, starting from the condition that gravitational force provides the centripetal force.(b) Calculate the orbital speed.(c) Calculate the orbital period in hours, and identify what type of orbit this is.

Show answer
(a) $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow v^2 = \dfrac{GM}{r} \Rightarrow v = \sqrt{\dfrac{GM}{r}}$(b) $v = \sqrt{\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{4.22\times10^{7}}} = 3.07\times10^{3}\,\text{m s}^{-1}$(c) $T = \dfrac{2\pi r}{v} = 8.63\times10^{4}\,\text{s} = 24.0\,\text{hours}$; this is a geostationary orbit.
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3

An $800\,\text{kg}$ satellite is raised from a circular orbit of altitude $300\,\text{km}$ to one of altitude $800\,\text{km}$ above the Earth’s surface (radius $6.37\times10^{6}\,\text{m}$).(a) State the formula for gravitational potential energy of a mass in a radial field.(b) Calculate the minimum work required to move the satellite between these orbits, and explain why your answer is positive.

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(a) $E_p = -\dfrac{GMm}{r}$(b) With $r_1 = 6.67\times10^{6}\,\text{m}$ and $r_2 = 7.17\times10^{6}\,\text{m}$, $W = E_{p2} – E_{p1} = 3.33\times10^{9}\,\text{J}$. It is positive because work must be done against the attractive gravitational force to move the satellite further from Earth.
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4

Explain, in terms of the gravitational potential gradient, why the gravitational field lines around a planet are radial and why the surfaces of constant potential are spheres centred on the planet.

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The field is spherically symmetric so field strength depends only on distance $r$ from the centre; field lines point radially inward (direction of steepest potential decrease). Equipotential surfaces are perpendicular to field lines, so they are spheres of constant $r$, and no work is done moving along one.
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Section B – Electric fields

5

Two parallel plates $2.0\,\text{cm}$ apart have a potential difference of $500\,\text{V}$ across them.(a) Calculate the electric field strength between the plates.(b) Calculate the force on an electron placed between the plates.(c) Calculate the electron’s acceleration (electron mass $= 9.11\times10^{-31}\,\text{kg}$).

Show answer
(a) $E = \dfrac{V}{d} = \dfrac{500}{0.020} = 2.5\times10^{4}\,\text{V m}^{-1}$(b) $F = eE = (1.60\times10^{-19})(2.5\times10^{4}) = 4.0\times10^{-15}\,\text{N}$(c) $a = \dfrac{F}{m} = \dfrac{4.0\times10^{-15}}{9.11\times10^{-31}} = 4.4\times10^{15}\,\text{m s}^{-2}$
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6

Two point charges, $+3.0\,\mu\text{C}$ and $-5.0\,\mu\text{C}$, are separated by $15\,\text{cm}$ in air. Calculate the magnitude of the electrostatic force between them and state whether it is attractive or repulsive.

Show answer
$F = \dfrac{k q_1 q_2}{r^2} = \dfrac{(8.99\times10^{9})(3.0\times10^{-6})(5.0\times10^{-6})}{(0.15)^2} = 6.0\,\text{N}$ (3 s.f.), attractive since the charges have opposite signs.
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7

A point charge of $+2.0\,\text{nC}$ is fixed in space.(a) Calculate the electric field strength it produces at a distance of $0.30\,\text{m}$.(b) Sketch-by-description how the field strength varies with distance from the charge, and state the name of this relationship.

Show answer
(a) $E = \dfrac{kq}{r^2} = \dfrac{(8.99\times10^{9})(2.0\times10^{-9})}{(0.30)^2} = 2.0\times10^{2}\,\text{V m}^{-1}$(b) Field strength falls smoothly with distance, decreasing steeply close to the charge and levelling off further away, following $E \propto \dfrac{1}{r^2}$ (an inverse square law).
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8

Compare a uniform electric field between two charged parallel plates with the radial gravitational field around the Earth, in terms of (a) the shape of the field lines and (b) whether the field strength depends on distance.

Show answer
(a) The electric field between the plates is uniform with straight, equally spaced parallel field lines; the gravitational field around the Earth is radial with lines converging on the centre.(b) The uniform electric field strength is constant (independent of position between the plates); the gravitational field strength decreases with distance following an inverse square law, $g \propto \dfrac{1}{r^2}$.
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Section C – Magnetic fields and electromagnetic induction

Exam-style and extended questions
9

A straight wire of length $12\,\text{cm}$ carries a current of $4.0\,\text{A}$ and lies at right angles to a uniform magnetic field of flux density $0.25\,\text{T}$.(a) Calculate the force on the wire.(b) The wire is now turned so that it makes an angle of $30^{\circ}$ with the field direction, with the current unchanged. Calculate the new force on the wire.

Show answer
(a) $F = BIL = (0.25)(4.0)(0.12) = 0.12\,\text{N}$(b) $F = BIL\sin\theta = (0.25)(4.0)(0.12)\sin30^{\circ} = 0.060\,\text{N}$
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10

A proton (mass $1.67\times10^{-27}\,\text{kg}$, charge $1.60\times10^{-19}\,\text{C}$) travels at $2.5\times10^{6}\,\text{m s}^{-1}$ at right angles into a uniform magnetic field of flux density $0.80\,\text{T}$.(a) Explain why the proton moves in a circular path.(b) Calculate the radius of the circular path.

Show answer
(a) The magnetic force on the proton, $F = qvB$, is always perpendicular to its velocity, so it acts as a centripetal force, giving circular motion at constant speed.(b) $r = \dfrac{mv}{qB} = \dfrac{(1.67\times10^{-27})(2.5\times10^{6})}{(1.60\times10^{-19})(0.80)} = 3.3\times10^{-2}\,\text{m}$
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11

A straight conducting rod of length $0.50\,\text{m}$ moves at a constant velocity of $3.0\,\text{m s}^{-1}$ at right angles to a uniform magnetic field of flux density $0.40\,\text{T}$, with the rod perpendicular to both its velocity and the field.(a) State the equation for the e.m.f. induced in the rod.(b) Calculate the e.m.f. induced.

Show answer
(a) $\varepsilon = BLv$(b) $\varepsilon = (0.40)(0.50)(3.0) = 0.60\,\text{V}$
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12

A flat coil of $200$ turns and cross-sectional area $8.0\times10^{-3}\,\text{m}^2$ is placed with its plane perpendicular to a magnetic field. The flux density through the coil increases uniformly from $0.2\,\text{T}$ to $0.8\,\text{T}$ in $0.15\,\text{s}$.(a) Calculate the magnitude of the e.m.f. induced in the coil.(b) State Lenz’s law and use it to explain the direction of the induced current in the coil.

Show answer
(a) $\varepsilon = N\dfrac{\Delta\Phi}{\Delta t} = N A\dfrac{\Delta B}{\Delta t} = (200)(8.0\times10^{-3})\dfrac{(0.8-0.2)}{0.15} = 6.4\,\text{V}$(b) Lenz’s law: the direction of an induced e.m.f./current is such that it opposes the change producing it. Since the flux through the coil is increasing, the induced current flows in the direction that creates a magnetic field opposing this increase (i.e. opposing the applied field), by conservation of energy.
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