Question

A charged particle is moving in a uniform magnetic field penetrates a layer of lead and thereby loss half of its kinetic energy. Then the radius of curvature of its path after loss in its kinetic energy will be

• A. Same as its initial value
• B. Reduced to $\frac{1}{2}$ times of its initial values
• C. Reduced to $\frac{1}{\sqrt{2}}$ times of its initial values
• D. Reduced to $\frac{1}{4}$ times of its initial values

Answer 1

Answer: (C)

Reduced to $\frac{1}{\sqrt{2}}$ times of its initial values

Answer 2

When a charged particle moves in a uniform magnetic field perpendicular to its velocity, it follows a circular path. The magnetic force provides the necessary centripetal force:

$qvB = \frac{mv^2}{r}$

From this, the radius of the circular path is $r = \frac{mv}{qB}$.

The kinetic energy of the particle is $K = \frac{1}{2}mv^2$. From the kinetic energy, the speed $v$ can be expressed as $v = \sqrt{\frac{2K}{m}}$.

Substitute this expression for $v$ into the radius formula:

$r = \frac{m}{qB} \sqrt{\frac{2K}{m}} = \frac{1}{qB} \sqrt{\frac{2m^2K}{m}} = \frac{1}{qB} \sqrt{2mK}$.

Let the initial kinetic energy be $K_0$ and the initial radius be $r_0$. $r_0 = \frac{1}{qB} \sqrt{2mK_0}$.

After penetrating the layer of lead, the particle loses half of its kinetic energy. So, the final kinetic energy $K_f$ is: $K_f = K_0 - \frac{1}{2}K_0 = \frac{1}{2}K_0$.

Let the final radius be $r_f$. $r_f = \frac{1}{qB} \sqrt{2mK_f} = \frac{1}{qB} \sqrt{2m\left(\frac{1}{2}K_0\right)} = \frac{1}{qB} \sqrt{mK_0}$.

To find the relationship between $r_f$ and $r_0$: $r_f = \frac{1}{qB} \sqrt{mK_0} = \frac{1}{qB} \sqrt{\frac{1}{2} (2mK_0)} = \frac{1}{\sqrt{2}} \left( \frac{1}{qB} \sqrt{2mK_0} \right)$.

Since $r_0 = \frac{1}{qB} \sqrt{2mK_0}$, we have: $r_f = \frac{1}{\sqrt{2}} r_0$.

Therefore, the radius of curvature of its path is reduced to $\frac{1}{\sqrt{2}}$ times its initial value.