# 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.

Demonstration

CRT Magic

Magnetic Pump

Concept Test

2836

[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.