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How does a magnetic field exert a force on a current-carrying conductor and on a moving charge?

Define magnetic flux density, calculate the force on a current-carrying conductor and on a moving charge, and analyse the circular motion of a charge in a magnetic field

A focused answer to the H2 Physics learning outcome on magnetic forces. Magnetic flux density, the force F = BIL on a current, the force F = Bqv on a moving charge, Fleming's left-hand rule, and circular motion in a field.

Generated by Claude Opus 4.89 min answer

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  1. What this dot point is asking
  2. The answer
  3. Examples in context
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What this dot point is asking

SEAB wants you to define magnetic flux density, calculate the force on a current-carrying conductor and on a moving charge, use Fleming's left-hand rule for direction, and analyse the circular motion of a charged particle in a uniform field. This is the motor-effect half of electromagnetism.

The answer

Magnetic flux density

The magnetic flux density BB (the strength of a magnetic field) is defined through the force it exerts. For a straight wire of length LL carrying current II at angle θ\theta to a uniform field:

F=BILsinθF = BIL\sin\theta

When the current is perpendicular to the field (θ=90\theta = 90^\circ), the force is maximum, F=BILF = BIL. When the current is parallel to the field (θ=0\theta = 0), the force is zero. The unit of BB is the tesla (T), where 1 T=1 N A1m11\ \text{T} = 1\ \text{N A}^{-1}\text{m}^{-1}.

The direction: Fleming's left-hand rule

The force is always perpendicular to both the current and the field. Fleming's left-hand rule gives the direction: with the thumb, first finger and second finger mutually perpendicular, the First finger points along the Field, the seCond finger along the Current, and the thuMb gives the Motion (force).

Force on a moving charge

A single charge qq moving at velocity vv through a field BB experiences:

F=BqvsinθF = Bqv\sin\theta

maximum when the velocity is perpendicular to the field. Because the force is always perpendicular to the velocity, it does no work: it changes the direction of motion but not the speed.

Circular motion in a uniform field

A charge moving perpendicular to a uniform field follows a circular path, because the constant-magnitude force is always perpendicular to the velocity. The magnetic force provides the centripetal force:

Bqv=mv2r    r=mvBqBqv = \frac{mv^2}{r} \implies r = \frac{mv}{Bq}

So heavier or faster particles follow larger circles, while stronger fields or larger charges tighten the path. This is the principle behind mass spectrometers and cyclotrons.

Examples in context

Example 1. The electric motor. In a motor, a current-carrying coil sits in a magnetic field. The force F=BILF = BIL on the two sides of the coil acts in opposite directions, producing a turning couple that rotates the coil. A commutator reverses the current every half turn to keep the rotation going, the direct application of the motor effect.

Example 2. The mass spectrometer. Ions accelerated to a known speed enter a magnetic field and follow circular paths with radius r=mvBqr = \dfrac{mv}{Bq}. Since heavier ions curve less, measuring the radius separates ions by mass-to-charge ratio, letting chemists identify isotopes and compounds.

Try this

Q1. State the factors that determine the size of the force on a current-carrying conductor in a magnetic field. [2 marks]

  • Cue. F=BILsinθF = BIL\sin\theta: flux density BB, current II, length LL in the field, and the angle θ\theta between current and field.

Q2. A wire carries 6.0 A6.0\ \text{A} over a length 0.20 m0.20\ \text{m} perpendicular to a 0.40 T0.40\ \text{T} field. Find the force on it. [2 marks]

  • Cue. F=BIL=0.40×6.0×0.20=0.48 NF = BIL = 0.40 \times 6.0 \times 0.20 = 0.48\ \text{N}.

Q3. Explain why a charged particle moving in a uniform magnetic field travels in a circle at constant speed. [3 marks]

  • Cue. The force BqvBqv is always perpendicular to the velocity, so it does no work (constant speed) and provides a constant-magnitude centripetal force, giving a circular path.

Exam-style practice questions

Practice questions written in the style of SEAB exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Original5 marksA straight wire of length 0.15 m0.15\ \text{m} carries a current of 4.0 A4.0\ \text{A} at right angles to a uniform magnetic field of flux density 0.30 T0.30\ \text{T}. (a) Find the force on the wire. (b) State the direction relative to the field and current. (c) State how the force changes if the wire is rotated to lie along the field.
Show worked answer →

(a) Force on a current: F=BIL=0.30×4.0×0.15=0.18 NF = BIL = 0.30 \times 4.0 \times 0.15 = 0.18\ \text{N} (the wire is perpendicular to the field, so sinθ=1\sin\theta = 1).

(b) The force is perpendicular to both the current and the field, in the direction given by Fleming's left-hand rule.

(c) If the wire lies along the field, the angle between the current and field is zero, so F=BILsin0=0F = BIL\sin 0 = 0: there is no force.

Markers reward F=BILF = BIL for the perpendicular case, the force direction perpendicular to both current and field (Fleming's left-hand rule), and zero force when the current is parallel to the field.

Original5 marksA proton (charge +1.60×1019 C+1.60 \times 10^{-19}\ \text{C}, mass 1.67×1027 kg1.67 \times 10^{-27}\ \text{kg}) enters a uniform magnetic field of 0.40 T0.40\ \text{T} at right angles, moving at 2.0×106 m s12.0 \times 10^6\ \text{m s}^{-1}. (a) Find the magnetic force on it. (b) Find the radius of its circular path.
Show worked answer →

(a) Force on a moving charge: F=Bqv=0.40×1.60×1019×2.0×106=1.28×1013 NF = Bqv = 0.40 \times 1.60 \times 10^{-19} \times 2.0 \times 10^6 = 1.28 \times 10^{-13}\ \text{N}.

(b) The magnetic force provides the centripetal force: Bqv=mv2rBqv = \dfrac{mv^2}{r}, so r=mvBq=1.67×1027×2.0×1060.40×1.60×1019=3.34×10216.4×1020=0.052 mr = \dfrac{mv}{Bq} = \dfrac{1.67 \times 10^{-27} \times 2.0 \times 10^6}{0.40 \times 1.60 \times 10^{-19}} = \dfrac{3.34 \times 10^{-21}}{6.4 \times 10^{-20}} = 0.052\ \text{m}.

Markers reward F=BqvF = Bqv for the magnetic force, equating it to the centripetal force, and the radius r=mv/(Bq)r = mv/(Bq) with units.

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