Question

An electron of mass 9.1×10(-31)kg enters a 1.7T magnetic field at right angle to the field .if the initial energy of the electron is 4.3MeV, calculate the radius of rotation of the electron. Calculate also the centripetal force on the electron and the frequency of circulation

Answer 1

For the electron entering magnetic field, centripetal force is equal to Lorentz force.

$\displaystyle F= \frac{mv^2}{r} = qvB$

So,

$\displaystyle r = \frac{mv}{qB}$

The initial energy of the electron is 4.3MeV, which is a full energy $E = mc^2 + E_k = 0.511\; MeV + E_k =4.3 \; MeV$

$E_k = 3.789 \; MeV$

Let's calculate v with simultaneous conversion to SI

$\displaystyle v= \sqrt{\frac{2E_k}{m}} =\sqrt{\frac{2 \cdot 3.789 \cdot 1.6 \cdot 10^{-19}}{9.1 \cdot 10^{-31}}} = 1.15 \cdot 10^6 \; m/s$

Then

$\displaystyle r = \frac{9.1 \cdot 10^{-31} \cdot 1.15 \cdot 10^6}{1.6 \cdot 10^{-19} \cdot 1.7} =3.847 \cdot 10^{-6} \, m = 3.847\; \mu m$

The centripetal force on the electron:

$F =qvB = 1.6 \cdot 10^{-19} \cdot 1.15 \cdot 10^6 \cdot 1.7=3.128 \cdot 10^{-13} \; N$

The frequency of circulation:

$\displaystyle \omega =\frac{qB}{m} = \frac{1.6\cdot 10^{-19} \cdot 1.7}{9.1 \cdot 10^{-31}} = 0.3 \cdot 10^{12} \; rad/s$