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

Provided Information

Electron velocity vector: \( \vec{v} = (1.0 \times 10^7 \, \text{m/s}) \hat{i} + (0.5 \times 10^7 \, \text{m/s}) \hat{j} \)
Magnetic field vector: \( \vec{B} = (0.5 \, \text{mT}) \hat{j} = (0.5 \times 10^{-3} \, \text{T}) \hat{j} \)

Calculation of Magnetic Force
The magnetic force \( \vec{F} \) on a charged particle is calculated using \( \vec{F} = q (\vec{v} \times \vec{B}) \). For an electron, the charge \( q \) is \( -e \), where \( e \approx 1.6 \times 10^{-19} \, \text{C} \). The cross product \( \vec{v} \times \vec{B} \) is determined as follows:
\[ \vec{v} \times \vec{B} = (1.0 \times 10^7 \, \hat{i} + 0.5 \times 10^7 \, \hat{j}) \times (0.5 \times 10^{-3} \, \hat{j}) \]
Expanding the cross product:
\[ = (1.0 \times 10^7 \times 0.5 \times 10^{-3}) (\hat{i} \times \hat{j}) + (0.5 \times 10^7 \times 0.5 \times 10^{-3}) (\hat{j} \times \hat{j}) \]
Since \( \hat{i} \times \hat{j} = \hat{k} \) and \( \hat{j} \times \hat{j} = 0 \):
\[ = 5.0 \times 10^3 \, \hat{k} + 0 \]
\[ = 5.0 \times 10^3 \, \hat{k} \, \text{m/s} \cdot \text{T} \]
The resulting magnetic force on the electron is:
\[ \vec{F} = -e (5.0 \times 10^3 \, \hat{k}) = -5.0 \times 10^3 e \, \hat{k} \, \text{N} \]

Determining the Radius of the Circular Path
The magnetic force provides the centripetal force required for circular motion, given by \( F = \frac{mv^2}{r} \). Here, \( m \approx 9.11 \times 10^{-31} \, \text{kg} \) is the electron's mass, and \( v \) is the velocity component perpendicular to \( \vec{B} \), which is \( v_{\perp} = 1.0 \times 10^7 \, \text{m/s} \).
Equating the magnetic force magnitude to the centripetal force magnitude:
\[ e v_{\perp} B = \frac{m v_{\perp}^2}{r} \]
Solving for the radius \( r \):
\[ r = \frac{m v_{\perp}}{e B} \]
Substituting the given values:
\[ r = \frac{(9.11 \times 10^{-31} \, \text{kg}) \times (1.0 \times 10^7 \, \text{m/s})}{(1.6 \times 10^{-19} \, \text{C}) \times (0.5 \times 10^{-3} \, \text{T})} \]
\[ r = \frac{9.11 \times 10^{-24}}{0.8 \times 10^{-22}} \, \text{m} \]
\[ r = 0.1139 \, \text{m} \]

Calculating the Linear Distance Covered
The electron possesses a velocity component parallel to the magnetic field, \( v_{\parallel} = 0.5 \times 10^7 \, \text{m/s} \), which results in linear motion along the field direction.
The period \( T \) of one revolution is calculated using:
\[ T = \frac{2\pi r}{v_{\perp}} = \frac{2\pi \times 0.1139 \, \text{m}}{1.0 \times 10^7 \, \text{m/s}} \]
\[ T \approx 7.16 \times 10^{-8} \, \text{s} \]
The linear distance \( d \) covered during this period is:
\[ d = v_{\parallel} \times T = (0.5 \times 10^7 \, \text{m/s}) \times (7.16 \times 10^{-8} \, \text{s}) \]
\[ d \approx 0.358 \, \text{m} \]

Summary of Results
The calculated radius of the circular path is approximately \( 0.114 \, \text{m} \).
The electron follows a helical trajectory, covering a linear distance of approximately \( 0.358 \, \text{m} \) in one complete revolution.