Electron Motion in Magnetic Field

When an electron is shot into a uniform magnetic field at an angle normal to the field lines, it experiences a magnetic force. This magnetic force acts as the centripetal force, causing the electron to move in a circular orbit. We can determine the radius of this circular orbit and the frequency of the electron's revolution using fundamental physics principles.

Magnetic Field and Electron Speed

Let's list the given values for the electron and the magnetic field:

• Magnetic field strength, \(B = 4.5 \, G\). We need to convert this to Tesla (T) as the standard SI unit:
• \(1 \, G = 10^{-4} \, T\)
• So, \(B = 4.5 \times 10^{-4} \, T\)
• Speed of the electron, \(v = 4.8 \times 10^6 \, m/s\)
• Charge of the electron, \(e = q = 1.6 \times 10^{-19} \, C\)
• Mass of the electron, \(m_e = m = 9.1 \times 10^{-31} \, kg\)

Radius of Circular Orbit Calculation

For an electron moving perpendicular to a uniform magnetic field, the magnetic force provides the necessary centripetal force for circular motion.

The magnetic force (\(F_B\)) on a charged particle moving in a magnetic field is given by:

\(F_B = qvB\sin\theta\)

Since the electron is shot normal to the field, \(\theta = 90^\circ\), and \(\sin(90^\circ) = 1\). So, the magnetic force simplifies to:

\(F_B = qvB\)

The centripetal force (\(F_c\)) required for circular motion is given by:

\(F_c = \frac{mv^2}{r}\)

Where \(r\) is the radius of the circular orbit.

Equating the magnetic force to the centripetal force for stable circular motion:

\(qvB = \frac{mv^2}{r}\)

Now, we can rearrange the formula to solve for the radius (\(r\)):

\(r = \frac{mv^2}{qvB}\)

\(r = \frac{mv}{qB}\)

Substitute the given values into the formula:

\(r = \frac{(9.1 \times 10^{-31} \, kg) \times (4.8 \times 10^6 \, m/s)}{(1.6 \times 10^{-19} \, C) \times (4.5 \times 10^{-4} \, T)}\)

\(r = \frac{43.68 \times 10^{-31+6}}{7.2 \times 10^{-19-4}}\)

\(r = \frac{43.68 \times 10^{-25}}{7.2 \times 10^{-23}}\)

\(r = \left(\frac{43.68}{7.2}\right) \times 10^{-25 - (-23)}\)

\(r = 6.0666... \times 10^{-2} \, m\)

Converting meters to centimeters (since \(1 \, m = 100 \, cm\)):

\(r = 6.0666... \times 10^{-2} \times 100 \, cm\)

\(r = 6.0666... \, cm\)

Rounding to one decimal place, the radius of the circular orbit is approximately:

\(r \approx 6.0 \, cm\)

Frequency of Revolution Calculation

The frequency of revolution (\(f\)) is the number of complete orbits per second. It can be related to the angular frequency (\(\omega\)) or directly derived from the forces.

We know that the angular frequency \(\omega = \frac{v}{r}\). Also, \(\omega = 2\pi f\).

Therefore, \(f = \frac{v}{2\pi r}\).

Alternatively, from the force balance \(qvB = \frac{mv^2}{r}\), we can find the angular frequency:

\(\frac{v}{r} = \frac{qB}{m}\)

So, \(\omega = \frac{qB}{m}\)

Since \(f = \frac{\omega}{2\pi}\), we have:

\(f = \frac{qB}{2\pi m}\)

Substitute the given values into this formula:

\(f = \frac{(1.6 \times 10^{-19} \, C) \times (4.5 \times 10^{-4} \, T)}{2 \times 3.14159 \times (9.1 \times 10^{-31} \, kg)}\)

\(f = \frac{7.2 \times 10^{-23}}{57.17318 \times 10^{-31}}\)

\(f = \left(\frac{7.2}{57.17318}\right) \times 10^{-23 - (-31)}\)

\(f = 0.12592... \times 10^8 \, Hz\)

\(f = 12.592... \times 10^6 \, Hz\)

Since \(1 \, MHz = 10^6 \, Hz\), we can express the frequency in MHz:

\(f = 12.592... \, MHz\)

Rounding to one decimal place, the frequency of revolution is approximately:

\(f \approx 12.6 \, MHz\)

Summary of Results

The calculated approximate values are:

These values match one of the given options.