Question

In a chamber, a uniform magnetic field of $6.5 \,G \,(1 G = 10^ {–4 }T)$ is maintained. An electron is shot into the field with a speed of $4.8 × 10^6\, m s^{–1}$ normal to the field. Obtain the frequency of revolution of the electron in its circular orbit. Does the answer depend on the speed of the electron? Explain.$(e = 1.5 × 10^{–19}C,\, m_e = 9.1×10^{–31} kg)$

Answer 1

Magnetic field strength, $B = 6.5 × 10^{−4 }\,T$
Charge of the electron,$e = 1.6 × 10^{−19} \,C$
Mass of the electron, $m_e = 9.1 × 10^{−31} kg$
Velocity of the electron, $v = 4.8 × 10^6 ms^{-1}$
Radius of the orbit, $r = 4.2\,cm = 0.042\, m$
Frequency of revolution of the electron = $v$
Angular frequency of the electron $=ω = 2π$v
Velocity of the electron is related to the angular frequency as: $v =rω$
In the circular orbit, the magnetic force on the electron is balanced by the centripetal force. Hence, we can write:
$\frac{mv^2}{r} = evB$
$eB = \frac{mv}{r} = \frac{m(rω)}{r} = \frac{m(r.2πv)}{r}$
$v = \frac{Be}{ 2πm}$
This expression for frequency is independent of the speed of the electron. On substituting the known values in this expression, we get the frequency as:
$v = \frac{6.5 × 10^{-4} × 1.6 × 10{-19}}{ 2 × 3.14 × 9.1 × 10^{-31}} = 1.82 × 10^6 Hz ≈ 18\,MHz$
Hence, the frequency of the electron is around 18 MHz and is independent of the speed of the electron.