Question

An electron moving with a velocity \( \vec{v} = (1.0 \times 10^7 \, \text{m/s}) \hat{i} + (0.5 \times 10^7 \, \text{m/s}) \hat{j} \) enters a region of uniform magnetic field \( \vec{B} = (0.5 \, \text{mT}) \hat{j} \). 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.}

Hint:

When a charged particle moves in a magnetic field, it follows a circular path, and the radius of this path depends on the velocity, charge, and magnetic field. The linear distance covered during one revolution is the circumference of the circle.

Answer 1

Given Data

Velocity of the electron, \( \vec{v} = (1.0 \times 10^7 \, \text{m/s}) \hat{i} + (0.5 \times 10^7 \, \text{m/s}) \hat{j} \) 
Magnetic field, \( \vec{B} = (0.5 \, \text{mT}) \hat{j} = (0.5 \times 10^{-3} \, \text{T}) \hat{j} \)

 Magnetic Force
The magnetic force \( \vec{F} \) on a moving charge is given by: \[ \vec{F} = q (\vec{v} \times \vec{B}) \] For an electron, \( q = -e \), where \( e \) is the elementary charge (\( e \approx 1.6 \times 10^{-19} \, \text{C} \)). The cross product \( \vec{v} \times \vec{B} \) is:
\[ \vec{v} \times \vec{B} = (1.0 \times 10^7 \, \hat{i} + 0.5 \times 10^7 \, \hat{j}) \times (0.5 \times 10^{-3} \, \hat{j}) \] \[ = 1.0 \times 10^7 \times 0.5 \times 10^{-3} (\hat{i} \times \hat{j}) + 0.5 \times 10^7 \times 0.5 \times 10^{-3} (\hat{j} \times \hat{j}) \] \[ = 5.0 \times 10^3 (\hat{k}) + 0 \] \[ = 5.0 \times 10^3 \, \hat{k} \, \text{m/s} \cdot \text{T} \] Therefore, the force is:
\[ \vec{F} = -e (5.0 \times 10^3 \, \hat{k}) = -5.0 \times 10^3 e \, \hat{k} \, \text{N} \] 

Radius of Circular Path
The centripetal force required for circular motion is provided by the magnetic force:
\[ F = \frac{mv^2}{r} \] where \( m \) is the mass of the electron (\( m \approx 9.11 \times 10^{-31} \, \text{kg} \)) and \( v \) is the component of velocity perpendicular to \( \vec{B} \).
The perpendicular component of velocity is \( v_{\perp} = 1.0 \times 10^7 \, \text{m/s} \).
Equating the magnetic force to the centripetal force:
\[ e v_{\perp} B = \frac{m v_{\perp}^2}{r} \] \[ r = \frac{m v_{\perp}}{e B} \] \[ r = \frac{9.11 \times 10^{-31} \times 1.0 \times 10^7}{1.6 \times 10^{-19} \times 0.5 \times 10^{-3}} \] \[ r = \frac{9.11 \times 10^{-24}}{0.8 \times 10^{-22}} \] \[ r = 0.1139 \, \text{m} \] Linear Distance Covered
The electron also has a velocity component parallel to the magnetic field (\( v_{\parallel} = 0.5 \times 10^7 \, \text{m/s} \)), which causes it to move linearly along the field direction.
The time period \( T \) for one revolution is:
\[ T = \frac{2\pi r}{v_{\perp}} = \frac{2\pi \times 0.1139}{1.0 \times 10^7} \] \[ T \approx 7.16 \times 10^{-8} \, \text{s} \] The linear distance \( d \) covered during one revolution is: \[ d = v_{\parallel} \times T = 0.5 \times 10^7 \times 7.16 \times 10^{-8} \] \[ d \approx 0.358 \, \text{m} \] Final Answer 
The radius of the circular path is approximately \( 0.114 \, \text{m} \). 
The electron traces a helical path, and the linear distance covered during one revolution is approximately \( 0.358 \, \text{m} \).