Understanding the Physics: Charged Particle Motion

This question explores the relationship between the radius of a charged particle's circular path in a magnetic field and the potential difference through which it was accelerated, along with its final speed. We need to find out how the radius ($r$) is proportional to the potential difference ($V$) and the speed ($u$).

Relating Potential Difference and Speed

When a charged particle with charge '$q$' and mass '$m$' is accelerated through a potential difference '$V$', it gains kinetic energy. This gain in kinetic energy is equal to the work done on the particle by the electric field, which is given by '$qV$'.

Let the initial speed be zero and the final speed be '$u$'. The kinetic energy gained is $\frac{1}{2}mu^2$. Equating the work done and the kinetic energy gained:

$ \frac{1}{2}mu^2 = qV $

From this equation, we can express the speed '$u$' in terms of the potential difference '$V$':

$ u^2 = \frac{2qV}{m} \implies u = \sqrt{\frac{2qV}{m}} $

This shows that the speed '$u$' is proportional to the square root of the potential difference '$V$', or $u \propto \sqrt{V}$. Consequently, $u^2 \propto V$.

Calculating Radius in a Magnetic Field

When the charged particle enters a uniform magnetic field '$B$' perpendicularly, it experiences a magnetic force ($F_B$). This force acts as the centripetal force ($F_c$), causing the particle to move in a circular path of radius '$r$'.

The magnetic force is given by:

$ F_B = q u B $

The centripetal force required for circular motion is:

$ F_c = \frac{mu^2}{r} $

Equating these two forces:

$ q u B = \frac{mu^2}{r} $

Now, we can solve for the radius '$r$':

$ r = \frac{mu^2}{q u B} = \frac{mu}{qB} $

This equation shows that the radius '$r$' is directly proportional to the momentum ($mu$) of the particle and inversely proportional to the product of the charge ($q$) and the magnetic field strength ($B$).

Determining Proportionality to V and u

We have established that $r \propto u$ (since $m, q, B$ are constants for a given particle and field). We also know that $u \propto \sqrt{V}$, which implies $V \propto u^2$.

The question asks for the proportionality of '$r$' in terms of the given options, which involve ratios of '$V$' and '$u$'. Let's examine the ratio $V/u$:

$ \frac{V}{u} $

Substitute the relationship $V \propto u^2$ into this ratio:

$ \frac{V}{u} \propto \frac{u^2}{u} = u $

Since we found that $r \propto u$, it follows that:

$ r \propto \frac{V}{u} $

Thus, the radius of the circular path is proportional to the ratio $V/u$. Let's confirm this by substituting $u = \sqrt{\frac{2qV}{m}}$ into the expression for $r$: $ r = \frac{m}{qB} u = \frac{m}{qB} \sqrt{\frac{2qV}{m}} $ Now consider the ratio $\frac{V}{u}$: $ \frac{V}{u} = \frac{V}{\sqrt{\frac{2qV}{m}}} = V \sqrt{\frac{m}{2qV}} = \sqrt{V^2 \frac{m}{2qV}} = \sqrt{\frac{mV}{2q}} $ Comparing $r$ and $\frac{V}{u}$: $ r = \frac{m}{qB} \sqrt{\frac{2qV}{m}} = \frac{m}{\sqrt{m}} \frac{1}{q} \sqrt{\frac{2q}{B^2}} \sqrt{V} = \sqrt{\frac{m}{q}} \sqrt{\frac{2}{B^2}} \sqrt{V} $ $ \frac{V}{u} = \sqrt{\frac{m}{2q}} \sqrt{V} $ We can see that $r$ is proportional to $\sqrt{V}$ and $\frac{V}{u}$ is also proportional to $\sqrt{V}$. Let's check the expression $r = \frac{mu}{qB}$. We found $\frac{V}{u} = \frac{mu}{2q}$. Rearranging this, $mu = 2q \frac{V}{u}$. Substitute this expression for $mu$ into the equation for $r$: $ r = \frac{(2q \frac{V}{u})}{qB} = \frac{2}{B} \frac{V}{u} $ Since $\frac{2}{B}$ is a constant, this confirms $r \propto \frac{V}{u}$.

Conclusion

The radius '$r$' of the circular path followed by the charged particle is proportional to the ratio of the potential difference '$V$' to the speed '$u$'.