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: Express radius of curvature in terms of kinetic energy.
For a charged particle moving perpendicular to a magnetic field, the radius is:
\[ r = \frac{mv}{qB} \]

Step 2: Connect velocity to kinetic energy.
Kinetic energy $K = \frac{1}{2}mv^2$, so $v = \sqrt{\frac{2K}{m}}$. Substituting:
\[ r = \frac{m}{qB}\sqrt{\frac{2K}{m}} = \frac{\sqrt{2mK}}{qB} \] This shows that $r \propto \sqrt{K}$.

Step 3: Write the ratio of final to initial radius.
\[ \frac{r_f}{r_i} = \sqrt{\frac{K_f}{K_i}} \]

Step 4: Use the given condition.
The particle loses half its kinetic energy, so $K_f = K_i / 2$:
\[ \frac{r_f}{r_i} = \sqrt{\frac{K_i/2}{K_i}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}} \]

Step 5: Interpret the result.
The radius decreases by a factor of $1/\sqrt{2} \approx 0.707$. The particle curves more tightly in the magnetic field after losing energy.

Step 6: State the answer.
\[ \boxed{r_f = \frac{r_i}{\sqrt{2}}} \] The radius is reduced to $\dfrac{1}{\sqrt{2}}$ times the initial value.