Physics: Fields (Gravitational, Electric, Magnetic)
Fields (gravitational, electric, magnetic) is the A-level Physics topic that completes the study of field theory begun with gravitational and electric fields, by covering magnetic fields and electromagnetic induction, and by drawing the three field types together synoptically. It covers the force on a current-carrying conductor and on a moving charge in a magnetic field, the circular motion of charged particles in a magnetic field, and Faraday's and Lenz's laws of electromagnetic induction, including flux linkage and the transformer equation. AQA regularly examines it synoptically, comparing the vector nature, force laws and field patterns of gravitational, electric and magnetic fields in a single question.
Before you start
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Method
- Calculate the force on a current-carrying conductor in a magnetic field using F = B x I x L x sin(theta), where theta is the angle between the conductor and the field; this force is maximum (F = BIL) when the conductor is perpendicular to the field, and zero when it is parallel to the field.
- Calculate the force on a single moving charge in a magnetic field using F = B x Q x v x sin(theta), and use Fleming's left-hand rule to find its direction: First finger = Field, seCond finger = Current (or direction of positive charge flow), thuMb = Motion (force).
- For a charged particle moving in a uniform magnetic field with its velocity perpendicular to the field, recognise that the magnetic force always acts at right angles to the velocity, so it provides a centripetal force and the particle moves in a circle; equate the two force expressions, B x Q x v = m x v^2 / r, and rearrange to find the radius, r = m x v / (B x Q).
- For electromagnetic induction, calculate magnetic flux as phi = B x A (for a coil face held perpendicular to the field), and flux linkage as N x phi, where N is the number of turns on the coil; use Faraday's law, EMF = -(rate of change of flux linkage) = -(delta(N x phi) / delta t), to calculate an induced EMF.
- Apply Lenz's law to find the direction (not just the size) of an induced EMF or current: the induced current always flows in the direction that opposes the change that produced it (for example, opposing the motion of a magnet, or opposing an increasing flux), which is why a minus sign appears in Faraday's law.
- For a transformer, use the transformer equation, Vp / Vs = Np / Ns (primary voltage over secondary voltage equals primary turns over secondary turns), and for an ideal (100% efficient) transformer, combine this with Vp x Ip = Vs x Is (power in = power out) to relate current to the turns ratio as well.
- For synoptic 'compare the fields' questions, keep three distinctions straight: gravitational and electric fields obey an inverse-square force law for a point mass/charge while the magnetic field around a straight current-carrying wire does not; gravitational fields are always attractive while electric and magnetic forces can be attractive or repulsive; and the force on a moving charge in a magnetic field depends on its velocity and acts perpendicular to that velocity, unlike the gravitational or electric force on the same particle.
Worked example
A proton (mass m = 1.67 x 10^-27 kg, charge Q = 1.60 x 10^-19 C) enters a uniform magnetic field of flux density 0.500 T, travelling at a constant speed of 2.00 x 10^6 m/s in a direction perpendicular to the field. Show that the proton moves in a circular path, and calculate the radius of this circular path.
- Since the proton's velocity is perpendicular to the magnetic field, the magnetic force on it has magnitude F = B x Q x v, and by Fleming's left-hand rule this force always acts at right angles to the proton's velocity.
- A force that is constant in magnitude and always perpendicular to the velocity is precisely the condition for circular motion, so the proton moves in a circle, with the magnetic force providing the centripetal force.
- Equate the magnetic force to the centripetal force expression: B x Q x v = m x v^2 / r.
- Cancel one factor of v from each side and rearrange to make r the subject: r = m x v / (B x Q).
- Substitute the values: r = (1.67 x 10^-27 x 2.00 x 10^6) / (0.500 x 1.60 x 10^-19) = (3.34 x 10^-21) / (8.00 x 10^-20).
- Final answer: r = 4.18 x 10^-2 m (3 s.f.), or 4.18 cm.
Practice questions
Try each question, then tap to reveal the answer.
Q1State the equation for the force on a straight current-carrying conductor of length L, carrying current I, placed at right angles to a magnetic field of flux density B.Show answer
Answer: F = B x I x L.
Q2A wire of length 0.40 m carrying a current of 3.0 A is placed at right angles to a magnetic field of flux density 0.25 T. Calculate the force on the wire.Show answer
Answer: F = B x I x L = 0.25 x 3.0 x 0.40 = 0.30 N.
Q3An electron moves to the right through a magnetic field that points into the page. Using Fleming's left-hand rule, and taking care with the electron's negative charge, state the direction of the magnetic force on the electron: up the page or down the page.Show answer
Answer: Down the page. Fleming's left-hand rule applied to a positive charge moving right through a field into the page gives a force up the page; since the electron is negatively charged, the actual force on it is reversed, so it acts down the page.
Q4State the two conditions needed for an EMF to be induced in a coil, in terms of magnetic flux.Show answer
Answer: There must be a magnetic flux linking the coil, and this flux linkage must be changing over time; a steady, unchanging flux induces no EMF.
Q5A coil of 200 turns and cross-sectional area 0.0050 m^2 lies with its plane perpendicular to a uniform magnetic field. The field increases steadily from 0.020 T to 0.060 T in 0.50 s. Calculate the average EMF induced in the coil.Show answer
Answer: Flux linkage change = N x (delta B) x A = 200 x (0.060 - 0.020) x 0.0050 = 200 x 0.040 x 0.0050 = 0.040 Wb turns. EMF = flux linkage change / time = 0.040 / 0.50 = 0.080 V.
Q6A step-down transformer has 1000 turns on its primary coil and 100 turns on its secondary coil. The primary voltage is 230 V. Calculate the secondary voltage, assuming the transformer is ideal.Show answer
Answer: Vs = Vp x (Ns/Np) = 230 x (100/1000) = 23 V.
Q7State one similarity and one difference between the field patterns of a gravitational field around a point mass and an electric field around a point charge.Show answer
Answer: Similarity: both are radial fields that obey an inverse-square law (field strength proportional to 1/r^2). Difference: a gravitational field is always attractive, whereas an electric field can be attractive (towards a negative charge) or repulsive (away from a positive charge), depending on the sign of the charge creating it.
Q8Explain why a charged particle moving parallel to a magnetic field experiences no magnetic force.Show answer
Answer: The magnetic force is given by F = B x Q x v x sin(theta), where theta is the angle between the velocity and the field; when the particle moves parallel to the field, theta = 0 degrees and sin(0 degrees) = 0, so the force is zero.
Exam-style questions
Written in the style of a A Level Science exam paper, with a full mark scheme.
A bar magnet is pushed, north pole first, into a solenoid connected to a sensitive centre-zero ammeter, inducing a current. (a) State the polarity created at the end of the solenoid nearest the approaching magnet, and explain your answer using Lenz's law. (b) Explain, in terms of energy, why a force is needed to push the magnet into the solenoid, and identify the energy transfer taking place.
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A charged particle moving horizontally enters a region where a uniform electric field and a uniform magnetic field act simultaneously, at right angles to each other and to the particle's velocity, arranged so that the electric and magnetic forces on the particle act in opposite directions. The particle passes through undeflected. The electric field strength is 1.50 x 10^4 V/m and the magnetic flux density is 0.0300 T. Calculate the speed of the particle.
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A search coil is rotated at a constant rate inside a uniform magnetic field, and the EMF induced across the coil is displayed on an oscilloscope, producing a sinusoidal trace. (a) State what happens to the peak EMF induced if the rate of rotation of the coil is doubled, all other factors unchanged. (b) Explain your answer in terms of Faraday's law.
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Free printable worksheet
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