Animated Solution for Physics - Magnetic Effects of Current: A proton and an $\alpha$-particle (with their masses in the ratio of $1:4$ and charges in the ratio of $1:2$) are accelerated from rest through a potential difference $V$. If a uniform magnetic field $B$ is set up perpendicular to their velocities, the ratio of the radii $r_p:r_{\alpha }$ of the circular paths described by them will be

Imagine a charged particle, starting from rest, being accelerated through a potential difference. It gains speed and then shoots into a region with a uniform magnetic field. What happens next? It starts moving in a circular path! This is a classic scenario in physics that beautifully connects electrostatics with magnetism.

First, let's find out how much energy the particle gains. When a charge $q$ is accelerated through a potential difference $V$, the work done on it by the electric field converts entirely into its kinetic energy. By the Work-Energy Theorem, we can write:

Now, as it enters the magnetic field perpendicularly, it experiences a magnetic Lorentz force. This force is always perpendicular to the velocity vector, meaning it does no work and doesn't change the particle's speed. Instead, it provides the necessary centripetal force to bend the particle's path into a circle. The radius $r$ of this circular path is given by the famous formula:

We need to connect this radius to the accelerating potential. We know that momentum $p=mv$ is related to kinetic energy $K$ by the equation $p=\sqrt{2mK}$. Substituting $K=qV$, we get the momentum in terms of the potential:

Now, let's substitute this back into our radius formula:

Here is the catch! The problem states that both the proton and the alpha particle are accelerated through the same potential difference $V$ and enter the same magnetic field $B$. So, $V$ and $B$ are constants for both particles. This means the radius is directly proportional to the square root of mass over charge:

Let's set up the ratio for the proton and the alpha particle. Using our proportionality, the ratio of their radii, $\frac{r_p}{r_{\alpha }}$, will be the square root of the ratio of their masses multiplied by the inverse ratio of their charges:

Now, we just plug in the given values. The mass ratio of proton to alpha particle is $\frac{m_p}{m_{\alpha }}=\frac{1}{4}$. The charge ratio is $\frac{q_p}{q_{\alpha }}=\frac{1}{2}$, which means the inverse charge ratio is $\frac{q_{\alpha }}{q_p}=\frac{2}{1}$. Don't make a silly mistake here by putting $\frac{1}{2}$ instead of $\frac{2}{1}$!

Inside the square root, we have $\frac{1}{4}$ multiplied by $2$. This simplifies to $\frac{2}{4}$, which is $\frac{1}{2}$. Taking the square root gives us:

So, the ratio of their radii is $1:\sqrt{2}$. We solved it! But think about this: what if the question said they had the same kinetic energy instead of the same accelerating potential? Or the same momentum? How would the proportionality change? These are favorite variations for JEE, so keep them in mind!