An electron moving with a velocity enters a region of uniform magnetic field Vector V = (1.0 × 10^7 m/s)i

Question

An electron moving with a velocity \(\vec ν = (1.0\times 10^7 \ m/s) \hat i + (0.5 \times 10^7\ m/s)\hat j\) enters a region of uniform magnetic field \(\vec B = (0.5 \ m T)\hat j \). Find the radius of the circular path described by it. While rotating; does the electron trace a linear path too? If so, calculate the linear distance covered by it during the period of one revolution.

Answer 1

Given, \(\vec ν = (1.0\times 10^7\hat i) + (0.5\times 10^7 \hat j)\)

\(\vec B = (0.5\times 10^{-3} \hat j) \text {tesla}\)

\(m_e = 9.1\times 10^{-31} \ kg\)

\(qe = 1.6 \times 10^{-19} C\)

Radius of path, \(R = \frac{mv_{⊥}}{qB}\)

Where \(V_{⊥}\) is component of velocity perpendicular to magnetic field.

R = 11.375 × 10^–2 m = 11.375 cm

Since the electron has perpendicular and parallel components of velocity, it will move on helical path, hence it will trace a linear path too virtually. The linear distance covered by it during one period of revolution is equal to pitch (P)

\(P = V_{||}\times T\)         \(\left\{T= \frac{2\pi m}{qB}\right\}\)

\(P = \frac{0.5\times 10^7\times 2\pi\times9.1\times 10^{-31}}{1.6\times 10^{-19} \times 0.5\times 10^{-3}} = \frac{18.2 \pi \times 10^{-24}}{1.6\times 10^{-22}}\)

\(= 11.375 \pi \times 10^{-2} m \ \text{or} \ 11.375\pi \ cm\)