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. Kinetic Energy and Radius of the Path: 

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

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

In the presence of a uniform magnetic field, the charged particle moves in a circular path with a radius \( r \), given by:

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

Where:

• \( r \) is 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, and
• \( B \) is the magnetic field strength.

 

2. Effect of Losing Half the Kinetic Energy:

After passing through the sheet of lead, the particle loses half of its kinetic energy, so the new kinetic energy is:

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

Since kinetic energy is proportional to the square of the velocity, the new velocity \( v' \) of the particle is:

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

Now, the new radius \( r' \) of the particle’s path is given by the equation:

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

Substituting \( v' = \frac{v}{\sqrt{2}} \) into this equation, we get:

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

Conclusion for the Radius:

• The radius of the path decreases by a factor of \( \sqrt{2} \). So, the new radius \( r' = \frac{r}{\sqrt{2}} \).

3. Time Period of Revolution:

The time period \( T \) of revolution of the particle in a magnetic field is given by:

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

Since the velocity of the particle decreases by a factor of \( \sqrt{2} \), the new velocity is \( v' = \frac{v}{\sqrt{2}} \). Therefore, the new time period \( T' \) is:

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

Substituting \( v' = \frac{v}{\sqrt{2}} \) into the equation, we get:

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

Conclusion for the Time Period:

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

Final Summary:

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