An electron moves with a speed of 2.00 x 10\(^7\) ms\(^{-1}\) in an orbit in a uniform magnetic field of 1.20 x 10\(^{-3}\) T. Calculate the radius of the o...
Question 1 Report
An electron moves with a speed of 2.00 x 10\(^7\) ms\(^{-1}\) in an orbit in a uniform magnetic field of 1.20 x 10\(^{-3}\) T. Calculate the radius of the orbit. [Mass of an electron = 9.11 x 10\(^{-3}\) kg; charge on an electron = 1.61 x 10\(^{-19}\)C]
Answer Details
Principle. An electron moving at speed v perpendicular to a magnetic field B experiences a magnetic force \(qvB\) that provides the centripetal force for its circular orbit:
\[ qvB = \frac{mv^{2}}{r} \;\Rightarrow\; r = \frac{mv}{qB} \]Substituting \(v = 2.00\times10^{7}\ \text{m/s}\), \(B = 1.20\times10^{-3}\ \text{T}\), \(q = 1.6\times10^{-19}\ \text{C}\), and the electron mass \(m = 9.11\times10^{-31}\ \text{kg}\) (the printed \(10^{-3}\) is a misprint for \(10^{-31}\)):
\[ r = \frac{9.11\times10^{-31}\times2.00\times10^{7}}{1.6\times10^{-19}\times1.20\times10^{-3}} \] \[ r = \frac{1.822\times10^{-23}}{1.92\times10^{-22}} \approx 0.095\ \text{m} \]The radius of the orbit is about 0.095 m (9.5 cm).
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