A charge q is accelerated through a potential difference V . It is then passed normally through a uniform magnetic field , where it moves in a circle of radius r . The potential difference required to move it in a circle of radius 2 r is

Options

A. 2 V
B. 4 V
C. 1 V
D. 3 V

Answer

To solve the problem step by step, we can follow these steps:

### Step 1: Understand the relationship between potential difference, charge, and kinetic energy
When a charge \( q \) is accelerated through a potential difference \( V \), it gains kinetic energy given by:
\[ K.E. = qV \]

### Step 2: Relate kinetic energy to velocity
The kinetic energy can also be expressed in terms of mass \( m \) and velocity \( v \):
\[ K.E. = \frac{1}{2} mv^2 \]

### Step 3: Set the two expressions for kinetic energy equal
Equating the two expressions for kinetic energy, we have:
\[ qV = \frac{1}{2} mv^2 \]

### Step 4: Solve for velocity \( v \)
From the equation above, we can solve for \( v \):
\[ v^2 = \frac{2qV}{m} \]
Taking the square root gives:
\[ v = \sqrt{\frac{2qV}{m}} \]

### Step 5: Relate the magnetic force to centripetal force
When the charge moves in a magnetic field \( B \), the magnetic force provides the centripetal force required to keep the charge moving in a circular path of radius \( r \):
\[ qvB = \frac{mv^2}{r} \]

### Step 6: Substitute for \( v \) in the centripetal force equation
Substituting \( v \) from step 4 into the centripetal force equation:
\[ q \left(\sqrt{\frac{2qV}{m}}\right) B = \frac{m\left(\frac{2qV}{m}\right)}{r} \]

### Step 7: Simplify the equation
This simplifies to:
\[ qB\sqrt{\frac{2qV}{m}} = \frac{2qV}{r} \]
Cancelling \( q \) from both sides (assuming \( q \neq 0 \)):
\[ B\sqrt{\frac{2V}{m}} = \frac{2V}{r} \]

### Step 8: Rearranging the equation
Rearranging gives:
\[ \sqrt{V} = \frac{2VBr}{\sqrt{2m}} \]

### Step 9: Establish the relationship between potential difference and radius
From the above relation, we can see that:
\[ V \propto r^2 \]
This means if the radius is doubled (from \( r \) to \( 2r \)), the potential difference must increase by a factor of \( 4 \):
\[ V' = 4V \]

### Final Answer
Thus, the potential difference required to move the charge in a circle of radius \( 2r \) is:
\[ \boxed{4V} \]