14.3.2 Circular Motion in Magnetic Field

An electric charge in an electric field always experiences an electric force. A charge in a magnetic field, however, does not necessarily experience a magnetic force.

Stationary Charge

Just like a wire carrying zero current experiences no magnetic force, a stationary charge does not experience any magnetic force. This is because \displaystyle {{F}_{b}}=Bqv and \displaystyle v=0. In the absence of other forces, this charge simply remains at rest.

v parallel to B

Just like a current carrying conductor placed parallel to a magnetic field experiences no magnetic force, a charge moving parallel to the direction of B does not experience any magnetic force. This is because \displaystyle {{F}_{b}}={{B}_{\bot }}qv and \displaystyle \displaystyle {{B}_{\bot }}=0. In the absence of other forces, this charge simply continues moving at velocity v.

v perpendicular to B

A charge moving perpendicular to the direction of B, however, does experience the magnetic force \displaystyle \displaystyle {{F}_{b}}={{B}_{\bot }}qv. It is important you realize that Fb is always perpendicular to v no matter how v rotates and turns. If Fb is the only force acting on this charge, this charge will be traveling along a circular path![1]

For a particle of mass m and charge q, we have

\displaystyle \begin{aligned}({{F}_{{net}}}&=ma)\\Bqv&=m\frac{{{{v}^{2}}}}{r}\\r&=\frac{{mv}}{{Bq}}\end{aligned}

To find T, the time taken for one complete revolution, we can divide the distance by speed:

\displaystyle \begin{aligned}T&=\frac{{2\pi r}}{v}\\&=(2\pi \frac{{mv}}{{Bq}})\div v\\&=\frac{{2\pi m}}{{Bq}}\end{aligned}

Two interesting results:

  1. For the same v and B, the radius of circular motion is directly proportional to \displaystyle \frac{m}{q}, the mass to charge ratio. This relationship is the basis for mass spectroscopy.
  2. The time taken for a charged particle to complete one revolution is independent of its speed. This result is exploited in the design of particle accelerators called cyclotrons.

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[1] This is not so for the electrons drifting in a current-carrying conductor because they are constrained by the metallic bonds to drift inside the wire.

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