Question

A charged particle moving in a uniform magnetic field penetrates a layer of lead and thereby loses half of its kinetic energy, then the radius of curvature of its path is:

• A. No change
• B. Reduced by \(\dfrac{1}{2}\) times of its initial value
• C. Reduced to \(\dfrac{1}{\sqrt{2}}\) times of its initial value
• D. Reduced to \(\dfrac{1}{4}\) times of its initial value

Hint:

In a magnetic field, \[ r=\frac{mv}{qB}. \] If kinetic energy changes, first find the change in velocity using \[ K\propto v^2, \] then relate radius with velocity.

Answer 1

The Correct Option is C

Step 1: Recall the expression for radius in a magnetic field.
For a charged particle moving perpendicular to a magnetic field, \[ r=\frac{mv}{qB} \] Thus, \[ r\propto v \]

Step 2: Write the expression for kinetic energy.
Kinetic energy is \[ K=\frac{1}{2}mv^2 \] Hence, \[ K\propto v^2 \]

Step 3: Use the condition given in the question.
The particle loses half of its kinetic energy. Therefore, \[ K_f=\frac{K_i}{2} \] Using \[ K\propto v^2, \] we get \[ v_f^2=\frac{v_i^2}{2} \] Taking square root, \[ v_f=\frac{v_i}{\sqrt{2}} \]

Step 4: Find the new radius of curvature.
Since \[ r\propto v, \] therefore, \[ r_f=\frac{r_i}{\sqrt{2}} \]

Step 5: Final conclusion.
Hence, the radius of curvature becomes \[ \boxed{\frac{1}{\sqrt{2}}\text{ times its initial value}} \]