Question

In a cyclotron (one type of particle accelerator), a deuteron (of mass 2.00 u) reaches a final speed of 8.8% of the speed of light while moving in a circular path of radius 0.459 m. What magnitude of magnetic force is required to maintain the deuteron in a circular path?

Answer 1

For cyclotron, the magnetic force will provide the required centripetal force hence

$\frac{mv^2}{r}=qvB\Rightarrow B=\frac{mv}{qr}$

In this case the particle reaches the 8.8% of speed of light hence there will be relativistic effect i.e. mass will now be

$m=\frac{m_0}{\sqrt{1-\frac{v^2}{c^2}}}=\frac{m_0}{\sqrt{1-\frac{(0.088c)^2}{c^2}}}=1.0039m_0$

where $m_0$ is rest mass of particle.

now 1 u =1.66X10^(-27) kg and charge on deutron is 1.6 X 10^(-19) i.e.

$B=\frac{2\times1.0039\times 1.66\times10^{-27}\times 0.088 \times 3\times10^8}{0.459\times1.6\times10^{-19}}=1.9181 T$