Question

A charged particle is moving in a circular path with velocity \( \vec{V} \) in a uniform magnetic field \( \vec{B} \). It is made to pass through a sheet of lead and as a consequence, it loses one-half of its kinetic energy without change in its direction. How will:
1. the radius of its path change? 2. its time period of revolution change?

Hint:

When the kinetic energy of a charged particle decreases, its velocity decreases, which in turn reduces the radius of its circular path in a magnetic field.

Answer 1

1. Initial Kinetic Energy and Path Radius:

The kinetic energy \( K \) of a charged particle with mass \( m \) and velocity \( v \) is defined as:

\[ K = \frac{1}{2} m v^2 \]

When subjected to a uniform magnetic field, the particle follows a circular trajectory with radius \( r \), calculated as:

\[ r = \frac{mv}{qB} \]

Where:

• \( r \) represents the radius of the circular path.
• \( m \) is the mass of the particle.
• \( v \) is the velocity of the particle.
• \( q \) is the charge of the particle.
• \( B \) is the strength of the magnetic field.

2. Impact of Halving Kinetic Energy:

Upon passing through the lead sheet, the particle's kinetic energy is reduced by half. The resulting kinetic energy \( K' \) is:

\[ K' = \frac{1}{2} K = \frac{1}{2} \times \frac{1}{2} m v^2 = \frac{1}{4} m v^2 \]

Given that kinetic energy is proportional to the square of velocity, the particle's new velocity \( v' \) will be:

\[ v' = \frac{v}{\sqrt{2}} \]

The radius of the particle's new path, \( r' \), is determined by:

\[ r' = \frac{m v'}{q B} \]

Substituting the expression for \( v' \) into this equation yields:

\[ r' = \frac{m \frac{v}{\sqrt{2}}}{q B} = \frac{r}{\sqrt{2}} \]

Radius Conclusion:

• The path radius is reduced by a factor of \( \sqrt{2} \). The new radius is \( r' = \frac{r}{\sqrt{2}} \).

3. Time Period of Revolution:

The time period \( T \) for a particle's revolution in a magnetic field is given by:

\[ T = \frac{2 \pi m}{q B v} \]

As the particle's velocity decreases to \( v' = \frac{v}{\sqrt{2}} \), the new time period \( T' \) becomes:

\[ T' = \frac{2 \pi m}{q B v'} \]

Substituting \( v' \) into this equation results in:

\[ T' = \frac{2 \pi m}{q B \frac{v}{\sqrt{2}}} = \sqrt{2} \times \frac{2 \pi m}{q B v} = \sqrt{2} \times T \]

Time Period Conclusion:

• The time period of revolution increases by a factor of \( \sqrt{2} \). The new time period is \( T' = \sqrt{2} \times T \).

Overall Summary:

• The radius of the particle's path decreases by a factor of \( \sqrt{2} \).
• The time period of revolution increases by a factor of \( \sqrt{2} \).