Question

An electron of energy 45 eV is revolving in a circular path in magnetic field of intensity \(9 \times 10^{-5}\) weber/m². Find the radius of the circular path.

Hint:

The radius of the circular path of an electron in a magnetic field depends on its velocity, mass, and charge, as well as the strength of the magnetic field.

Answer 1

The kinetic energy \(K.E.\) of the electron is given as 45 eV. We can convert this into joules: \[ K.E. = 45 \, \text{eV} = 45 \times 1.602 \times 10^{-19} \, \text{J} = 7.209 \times 10^{-18} \, \text{J} \] The expression for the kinetic energy of an electron is also: \[ K.E. = \frac{1}{2} m v^2 \] Where: - \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron.
Rearranging for \(v\), we get: \[ v = \sqrt{\frac{2 K.E.}{m}} = \sqrt{\frac{2 \times 7.209 \times 10^{-18}}{9.11 \times 10^{-31}}} = 1.268 \times 10^7 \, \text{m/s} \] Now, using the formula for the radius of the circular path in a magnetic field: \[ r = \frac{mv}{qB} \] Where: - \(r\) is the radius,
- \(m\) is the mass of the electron,
- \(v\) is the velocity of the electron,
- \(q\) is the charge of the electron,
- \(B\) is the magnetic field strength.
Substituting the known values: - \(q = 1.602 \times 10^{-19} \, \text{C}\),
- \(B = 9 \times 10^{-5} \, \text{T}\),
We get: \[ r = \frac{9.11 \times 10^{-31} \times 1.268 \times 10^7}{1.602 \times 10^{-19} \times 9 \times 10^{-5}} = 0.079 \, \text{m} \] Thus, the radius of the circular path is \(0.079 \, \text{m}\).