Question

An electron moves at \[3 \times 10^7\] m/s in a circular path of radius 0.28 m in a magnetic field. Its frequency of revolution is approximately:

Options

• 17 GHz

• 34 MHz

• 17 MHz

• 1.7 MHz

Answer 1

The frequency is \[\nu = \frac{v}{2\pi r} = \frac{3 \times 10^7}{2\pi \times 0.28}\]. Computing: \[2\pi \times 0.28 \approx 1.76\] m, so \[\nu \approx \frac{3 \times 10^7}{1.76} \approx 1.7 \times 10^7\] Hz ≈ 17 MHz.