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A2 · Practice questions · Magnetic fields due to currents

Each wire sits in the other's field.

Six original Cambridge-style questions on the fields of a straight wire, a flat coil and a solenoid, the effect of an iron core, and the force between parallel currents.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.
Keep these straight

Like currents attract, unlike currents repel.

01
Analysis
[3 marks]

Two long straight parallel wires carry currents in the same direction. Explain, in terms of the magnetic field of one wire and the force on the other, why the wires attract each other, and explain why the forces on the two wires are equal in magnitude.

  • The first wire produces a field of concentric circles, which at the position of the second wire is perpendicular to that wire. ✓
  • The second wire carries a current across this field, so it experiences a force F = BIL; Fleming's left-hand rule gives the direction as towards the first wire (and by the same argument the first wire is pushed towards the second). ✓
  • The forces are equal and opposite because they are a Newton's third law pair; between the wires the two circular fields point in opposite senses and partly cancel, so each wire is pushed into the weaker field. ✓
02
Calculation
[3 marks]

The flux density at a distance d from a long straight wire carrying a current I is B = μ₀I / 2πd, where μ₀ = 4π × 10⁻⁷ H m⁻¹. Wire P carries 12 A and wire Q, parallel to it and 4.0 cm away, carries 8.0 A in the opposite direction. Calculate the flux density at Q due to P, the force per unit length on Q, and state whether the wires attract or repel.

  • B = μ₀I / 2πd = 4π × 10⁻⁷ × 12 / (2π × 0.040) = 6.0 × 10⁻⁵ T. ✓
  • F / L = BIQ = 6.0 × 10⁻⁵ × 8.0 = 4.8 × 10⁻⁴ N m⁻¹. ✓
  • The currents are in opposite directions, so the wires repel; the same force per unit length acts on P. ✓
03
Calculation
[2 marks]

Two parallel wires carrying 5.0 A and 3.0 A, 2.5 cm apart, experience a force per unit length of 2.0 × 10⁻⁴ N m⁻¹. The force per unit length is proportional to the product of the currents and inversely proportional to the separation. Calculate the force per unit length when both currents are doubled and the separation is halved.

  • Doubling each current multiplies the force by 2 × 2 = 4, and halving the separation doubles it again, a factor of 8 in all. ✓
  • F / L = 8 × 2.0 × 10⁻⁴ = 1.6 × 10⁻³ N m⁻¹. ✓
04
Analysis
[3 marks]

A solenoid is viewed from one end, X, and the current is seen to circulate anticlockwise. (a) State the polarity of end X. (b) Describe the magnetic field inside and outside the solenoid. (c) State and explain the effect of inserting a soft iron core, and why this is useful in an electromagnet.

  • (a) An anticlockwise current seen from X makes X a north pole. ✓
  • (b) Inside, the field is strong and almost uniform (parallel, equally spaced lines along the axis); outside, it is like the field of a bar magnet, with lines leaving X and entering the far end. ✓
  • (c) The core is magnetised by the solenoid's field and greatly increases the flux density without changing the field's shape; because soft iron loses its magnetism when the current stops, the electromagnet can be switched on and off. ✓
05
Calculation
[3 marks]

Two long straight parallel wires, 6.0 cm apart, each carry a current of 10 A in the same direction. Use B = μ₀I / 2πd, with μ₀ = 4π × 10⁻⁷ H m⁻¹, to find the resultant flux density at the point midway between the wires, and at a point on the same line 3.0 cm beyond one of the wires (that is, 9.0 cm from the other).

  • At the midpoint each wire gives B = 2 × 10⁻⁷ × 10 / 0.030 = 6.7 × 10⁻⁵ T, but for same-direction currents the two fields between the wires are in opposite senses, so the resultant is zero. ✓
  • At the outer point the near wire gives 6.67 × 10⁻⁵ T and the far wire gives 2 × 10⁻⁷ × 10 / 0.090 = 2.22 × 10⁻⁵ T, now in the same sense. ✓
  • Resultant = 6.67 × 10⁻⁵ + 2.22 × 10⁻⁵ = 8.9 × 10⁻⁵ T. ✓
06
Calculation
[4 marks]

A long straight horizontal wire carries a current of 25 A towards the north. The horizontal component of the Earth's magnetic field is 1.8 × 10⁻⁵ T towards the north. Take the field of the wire to be B = μ₀I / 2πd, with μ₀ = 4π × 10⁻⁷ H m⁻¹.

(a) Calculate the distance from the wire at which its field equals the horizontal component of the Earth's field. (b) A compass is placed directly above the wire at this distance. State the direction of the wire's field there and the direction in which the compass points. (c) State the direction of the compass when the current is reversed.

  • (a) d = μ₀I / 2πB = 2 × 10⁻⁷ × 25 / 1.8 × 10⁻⁵. ✓
  • d = 0.28 m. ✓
  • (b) By the right-hand grip rule (thumb north) the field above a northward current points east; the two equal fields, east and north, add to a resultant pointing north-east, so the compass points 45° east of north. ✓
  • (c) Reversing the current makes the wire's field point west, so the compass points north-west (45° west of north). ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.