A particle of mass m and having charge q is moving in a circular path of radius r in the presence of a uniform magnetic field. The particle breaks into smaller fragments X and Y of equal mass. The fragment X now has charge `q/3` and the fragment Y has charge `(2q)/3`. Immediately after the fragmentation, X comes to rest. The radius of the path of Y is now `r_(Y)`. The ratio `(r_(Y))/r` is ______________.

Explanation

To solve the problem step by step, we will analyze the motion of the charged particle before and after it breaks into fragments X and Y.

### Step 1: Understand the initial conditions
Initially, we have a particle of mass \( m \) and charge \( q \) moving in a circular path of radius \( r \) in a uniform magnetic field \( B \). The centripetal force required for circular motion is provided by the magnetic force.

### Step 2: Write the equation for the radius of the circular path
The magnetic force acting on the particle is given by:
\[ F = qvB \]
where \( v \) is the velocity of the particle. The centripetal force required for circular motion is:
\[ F_c = \frac{mv^2}{r} \]
Setting the magnetic force equal to the centripetal force, we have:
\[ qvB = \frac{mv^2}{r} \]
From this, we can derive the expression for the radius \( r \):
\[ r = \frac{mv}{qB} \]

### Step 3: Analyze the fragmentation
The particle breaks into two fragments, X and Y, each of mass \( \frac{m}{2} \). Fragment X has a charge of \( \frac{q}{3} \) and fragment Y has a charge of \( \frac{2q}{3} \). After fragmentation, fragment X comes to rest, which means its velocity \( v_X = 0 \).

### Step 4: Apply conservation of momentum
Since no external forces are acting on the system, the momentum before fragmentation must equal the momentum after fragmentation. Before fragmentation, the total momentum is:
\[ P_{initial} = mv \]
After fragmentation, the momentum is:
\[ P_{final} = \left(\frac{m}{2}\right)(0) + \left(\frac{m}{2}\right)v_Y = \frac{m}{2}v_Y \]
Setting the initial momentum equal to the final momentum gives:
\[ mv = \frac{m}{2}v_Y \]
Solving for \( v_Y \):
\[ v_Y = 2v \]

### Step 5: Calculate the radius of the path of fragment Y
Now we can find the radius \( r_Y \) for fragment Y. The magnetic force acting on fragment Y is given by:
\[ F_Y = \left(\frac{2q}{3}\right)v_Y B \]
The centripetal force for fragment Y is:
\[ F_{c,Y} = \frac{\frac{m}{2}v_Y^2}{r_Y} \]
Setting the magnetic force equal to the centripetal force:
\[ \left(\frac{2q}{3}\right)(2v)B = \frac{\frac{m}{2}(2v)^2}{r_Y} \]
Simplifying this:
\[ \frac{4qvB}{3} = \frac{2mv^2}{r_Y} \]
Rearranging gives:
\[ r_Y = \frac{2mv^2}{\frac{4qvB}{3}} = \frac{3mv^2}{2qvB} \]

### Step 6: Relate \( r_Y \) to \( r \)
From the expression for \( r \):
\[ r = \frac{mv}{qB} \]
We can substitute this into our expression for \( r_Y \):
\[ r_Y = \frac{3}{2} \left(\frac{mv}{qB}\right) = \frac{3}{2} r \]

### Step 7: Find the ratio \( \frac{r_Y}{r} \)
Thus, we find:
\[ \frac{r_Y}{r} = \frac{3}{2} \]

### Final Answer
The ratio \( \frac{r_Y}{r} \) is \( \frac{3}{2} \) or \( 1.5 \).

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