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When an electron enters a magnetic field perpendicularly, it experiences a magnetic force. This force is unique because it acts perpendicular to both the velocity of the electron and the magnetic field itself.
This perpendicular force changes the direction of the electron's motion but not its speed. The outcome is a path that forms a perfect circle.
The magnetic force on the electron can be described by the formula:

• F_m = qvB

This formula tells us that the force (\(F_m\)) depends on the charge of the electron (\(q\)), its velocity (\(v\)), and the magnetic field strength (\(B\)).
Because this force is always perpendicular to the electron’s path, it constantly changes the electron's direction without altering its speed, thereby making the electron move in a circular path.

The concept of centripetal force is essential to understanding the circular motion of electrons in a magnetic field. A centripetal force is any force that keeps a body moving with a uniform speed along a circular path.
For the electron moving in a magnetic field, the magnetic force provides the necessary centripetal force to sustain its circular motion. This is described by:

• F_c = \frac{m_{e}v^2}{r}

Where:

• \(F_c\) is the centripetal force.
• \(m_{e}\) is the mass of the electron.
• \(v\) is its velocity.
• \(r\) is the radius of the circular path.

When an electron moves at a constant speed in a circle,
the magnetic force (\(F_m\)) equals the centripetal force (\(F_c\)). These forces balance each other leading to the equation:

• qvB = \frac{m_{e}v^2}{r}

This equation shows how the magnetic effect causes the electron to follow a curved circular pathway.

The radius of the circular path is a crucial aspect that determines how an electron behaves in a magnetic field.
To find this radius (\(r\)), we rearrange the equation for centripetal force and magnetic force balance:

• r = \frac{m_{e}v}{qB}

This rearranged formula shows the factors affecting the radius:

• The mass of the electron (\(m_{e}\))
• Its velocity (\(v\))
• The charge of the electron (\(q\))
• The magnetic field strength (\(B\))

Given the problem's specific values:

• \(m_{e}=9.1 \times 10^{-31} \text{ kg}\)
• \(v=4.8 \times 10^{6} \text{ m/s}\)
• \(q=1.6 \times 10^{-19} \text{ C}\)
• \(B=6.5 \times 10^{-4} \text{ T}\)

Placing these into the formula gives the radius:\[r = \frac{(9.1 \times 10^{-31} \text{ kg})(4.8 \times 10^{6} \text{ m/s})}{(1.6 \times 10^{-19} \text{ C})(6.5 \times 10^{-4} \text{ T})} = 0.216 \text{ m}\]Understanding these factors helps to comprehend why the electron moves in a circular path with this specific radius.