Questions: A proton travels at 0.999 c in a direction perpendicular to a uniform magnetic field of 7.0 T. The rest mass of a proton is m0 = 1.672 × 10^-27 kg and the charge on the proton is 1.6 × 10^-19 C. Calculate the radius of the circular path of the electron. Give your answer to one significant figure.

Solution Steps

The proton is traveling at a speed of \(0.999c\), where \(c\) is the speed of light. We need to calculate the relativistic mass \(m\) using the formula: \[ m = \frac{m_0}{\sqrt{1 - \left(\frac{v}{c}\right)^2}} \] where \(m_0 = 1.672 \times 10^{-27} \, \text{kg}\) and \(v = 0.999c\).

\[ m = \frac{1.672 \times 10^{-27} \, \text{kg}}{\sqrt{1 - (0.999)^2}} = \frac{1.672 \times 10^{-27} \, \text{kg}}{\sqrt{1 - 0.998001}} = \frac{1.672 \times 10^{-27} \, \text{kg}}{\sqrt{0.001999}} \approx \frac{1.672 \times 10^{-27} \, \text{kg}}{0.0447} \approx 3.740 \times 10^{-26} \, \text{kg} \]

The radius \(r\) of the circular path of a charged particle moving perpendicular to a magnetic field is given by: \[ r = \frac{mv}{qB} \] where \(m\) is the relativistic mass, \(v = 0.999c\), \(q = 1.6 \times 10^{-19} \, \text{C}\), and \(B = 7.0 \, \text{T}\).

First, calculate the momentum \(p = mv\): \[ p = (3.740 \times 10^{-26} \, \text{kg})(0.999 \times 3.00 \times 10^8 \, \text{m/s}) \approx 1.121 \times 10^{-17} \, \text{kg} \cdot \text{m/s} \]

Now, calculate the radius: \[ r = \frac{p}{qB} = \frac{1.121 \times 10^{-17} \, \text{kg} \cdot \text{m/s}}{(1.6 \times 10^{-19} \, \text{C})(7.0 \, \text{T})} \approx \frac{1.121 \times 10^{-17}}{1.12 \times 10^{-18}} \approx 10 \, \text{m} \]

Final Answer