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122–123ElectricityStretch

Magnetic Fields and Forces

A current-carrying wire in a magnetic field feels a force, and so does a single charged particle moving through one. Both are the same effect — magnetic fields push on moving charge — and because the force is always perpendicular to the motion, it can bend a particle into a perfect circle.

40 min Topic 7 · Electric and Magnetic Fields

What you'll be able to do

  • Describe magnetic flux density and its unit, the tesla
  • Use F=BIlsin⁡θF = BIl\sin\theta and Fleming’s left-hand rule
  • Use F=Bqvsin⁡θF = Bqv\sin\theta for a moving charge
  • Explain and calculate the circular motion of charged particles in a field
1

Magnetic flux density

The strength of a magnetic field is its magnetic flux density\textbf{magnetic flux density} BB, measured in teslas. A field of 1 T exerts a force of 1 N on a 1 m wire carrying 1 A at right angles to the field.

Field lines run from north to south outside a magnet. Around a straight current-carrying wire they form concentric circles, with direction given by the right-hand grip rule.

Typical values span a wide range: the Earth’s field is about 5×10−55\times10^{-5} T, a fridge magnet around 0.01 T, an MRI scanner several tesla.

1 T=1 N A−1 m−11\ \mathrm{T} = 1\ \mathrm{N\,A^{-1}\,m^{-1}}
The tesla.

Tip — Magnetic fields exert forces on moving charges and currents, not on stationary charges. A charge at rest in a magnetic field feels nothing.

2

Force on a current-carrying wire

A wire of length ll carrying current II at angle θ\theta to a field BB experiences a force F=BIlsin⁡θF = BIl\sin\theta. The force is maximum when the wire is perpendicular to the field and zero when parallel.

The direction is perpendicular to both the wire and the field, given by Fleming’s left-hand rule\textbf{Fleming’s left-hand rule}: first finger for Field, second finger for Current, thumb for the resulting Motion (force).

This is the motor effect\textbf{motor effect}, the principle behind electric motors and loudspeakers.

F=BIlsin⁡θF = BIl\sin\theta
Force on a wire.
Example 1A 0.25 m wire carrying 4.0 A lies at 30∘30^{\circ} to a 0.12 T field. Find the force.
1F=0.12×4.0×0.25×sin⁡30∘F = 0.12 \times 4.0 \times 0.25 \times \sin30^{\circ}.
2=0.12×0.5= 0.12 \times 0.5.
Answer0.060 N0.060\ \mathrm{N}
3

Force on a moving charge

A current is a flow of charges, so each moving charge must feel a force. For a charge qq moving at speed vv at angle θ\theta to the field, F=Bqvsin⁡θF = Bqv\sin\theta.

It follows from F=BIlF = BIl: in time tt, charge qq travels l=vtl = vt and constitutes a current I=q/tI = q/t, so BIl=B(q/t)(vt)=BqvBIl = B(q/t)(vt) = Bqv.

The direction again comes from Fleming’s left-hand rule, using conventional current. For a negative charge such as an electron, the force is in the opposite\textbf{opposite} direction to that for a positive charge.

F=Bqvsin⁡θF = Bqv\sin\theta
Force on a moving charge.
Example 1A proton moves at 2.0×106 m s−12.0\times10^{6}\ \mathrm{m\,s^{-1}} perpendicular to a 0.40 T field. Find the force.
1F=0.40×1.60×10−19×2.0×106F = 0.40 \times 1.60\times10^{-19} \times 2.0\times10^{6}.
Answer1.3×10−13 N1.3\times10^{-13}\ \mathrm{N}

Tip — For electrons, point the second finger opposite to their direction of motion, since conventional current is the flow of positive charge.

4

Circular paths

The magnetic force on a charge is always perpendicular to its velocity, so it does no work and cannot change the speed — only the direction. A particle moving perpendicular to a uniform field therefore travels in a circle\textbf{circle}.

The magnetic force provides the centripetal force: Bqv=mv2/rBqv = mv^{2}/r, giving r=mvBq=pBqr = \dfrac{mv}{Bq} = \dfrac{p}{Bq}.

Faster or heavier particles curve less; stronger fields or larger charges curve them more. This is the principle of the mass spectrometer, which separates ions by their radius of curvature, and of the cyclotron, where the time for a half-circle is independent of speed.

r=mvBqr = \dfrac{mv}{Bq}
Radius of the circular path.
T=2πmBqT = \dfrac{2\pi m}{Bq}
Period — independent of speed.
Example 1An electron moves at 5.0×106 m s−15.0\times10^{6}\ \mathrm{m\,s^{-1}} perpendicular to a 2.0 mT field. Find the radius of its path.
1r=9.11×10−31×5.0×1062.0×10−3×1.60×10−19r = \dfrac{9.11\times10^{-31} \times 5.0\times10^{6}}{2.0\times10^{-3} \times 1.60\times10^{-19}}.
2=4.555×10−243.2×10−22= \dfrac{4.555\times10^{-24}}{3.2\times10^{-22}}.
Answerr=1.4×10−2 mr = 1.4\times10^{-2}\ \mathrm{m}, about 1.4 cm.

Equation recap

F=BIlsin⁡θF = BIl\sin\theta
Force on a wire.
F=Bqvsin⁡θF = Bqv\sin\theta
Force on a charge.
r=mv/Bqr = mv/Bq
Circular path radius.
T=2πm/BqT = 2\pi m/Bq
Period in a field.

Common mistakes to avoid

Saying a stationary charge feels a magnetic force.
Only moving charges (or currents) experience magnetic forces.
Claiming the magnetic force speeds a particle up.
It is perpendicular to velocity, so it changes direction only.
Using the right-hand rule for the motor effect.
Fleming’s left\textit{left}-hand rule gives the force.
Forgetting mT or µT conversions.
Convert to teslas before substituting.

Key takeaways

  • Magnetic flux density BB is measured in teslas.
  • A wire in a field feels F=BIlsin⁡θF = BIl\sin\theta, direction by Fleming’s left-hand rule.
  • A moving charge feels F=Bqvsin⁡θF = Bqv\sin\theta; negative charges are forced the opposite way.
  • The force is perpendicular to velocity, so it does no work.
  • Charges perpendicular to a uniform field move in circles of radius r=mv/Bqr = mv/Bq.

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