An electron enters a 0.01 T magnetic field at 2·10⁶ m/s perpendicular to the field. Find the period of revolution (m = 9.1·10⁻³¹ kg, q = 1.6·10⁻¹⁹ C) — Physics problems | PhysLabo
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Period of an electron moving in a magnetic field

Medium 11 grade Electricity / Magnetism

An electron of mass $m = 9.1 \cdot 10^{-31}$ kg and charge $q = 1.6 \cdot 10^{-19}$ C enters a magnetic field $B = 0.01$ T with speed $v = 2 \cdot 10^6$ m/s perpendicular to the induction lines. Find the period of revolution of the electron.

Given
  • $m = 9.1 \cdot 10^{-31}$ kg
  • $q = 1.6 \cdot 10^{-19}$ C
  • $B = 0.01$ T
  • $T = ?$
Visual demonstration
Hints
Hint 1 / 2
The period of revolution does not depend on the speed: $T = 2\pi m/(qB)$.
Hint 2 / 2
Combine the exponents when multiplying: $10^{-19} \cdot 10^{-2} = 10^{-21}$.
Solution

The radius and period of circular motion of a charge in a magnetic field:

$$T = \frac{2\pi m}{qB}$$

$$T = \frac{2\pi \cdot 9.1 \cdot 10^{-31}}{1.6 \cdot 10^{-19} \cdot 0.01} = \frac{5.72 \cdot 10^{-30}}{1.6 \cdot 10^{-21}} \approx 3.57 \cdot 10^{-9} \text{ s}$$

Answer

$T \approx 3.57 \cdot 10^{-9}$ s