A proton (mass m) accelerated by a potential difference V flies through a uniform transverse magnetic field B. The field occupies a region of space by width d. If `alpha` be the angle of deviation of proton from initial direction of motion , the value of sin `alpha` will be

Options

A. `B/2 sqrt((qd)/(mv))`
B. `B/d sqrt(q/(2mV))`
C. `Bd sqrt(q/(2mV))`
D. `qvsqrt((Bd)/(2m))`

Explanation

To find the value of \(\sin \alpha\) for a proton accelerated by a potential difference \(V\) and flying through a uniform transverse magnetic field \(B\), we can follow these steps:

### Step-by-Step Solution:

1. **Determine the Kinetic Energy:**
The kinetic energy (\(KE\)) gained by the proton when it is accelerated through a potential difference \(V\) is given by:
\[ KE = qV \]
where \(q\) is the charge of the proton.

2. **Relate Kinetic Energy to Velocity:**
The kinetic energy can also be expressed in terms of the mass (\(m\)) and velocity (\(v\)) of the proton:
\[ KE = \frac{1}{2} mv^2 \]
Setting the two expressions for kinetic energy equal gives:
\[ qV = \frac{1}{2} mv^2 \]

3. **Solve for Velocity:**
Rearranging the equation to solve for \(v\):
\[ v^2 = \frac{2qV}{m} \]
Taking the square root:
\[ v = \sqrt{\frac{2qV}{m}} \]

4. **Magnetic Force as Centripetal Force:**
When the proton enters the magnetic field, it experiences a magnetic force that acts as the centripetal force. The magnetic force (\(F_B\)) is given by:
\[ F_B = qvB \]
This force provides the centripetal force needed to keep the proton moving in a circular path:
\[ F_c = \frac{mv^2}{r} \]
Setting the magnetic force equal to the centripetal force:
\[ qvB = \frac{mv^2}{r} \]

5. **Substituting for Velocity:**
Substitute \(v\) from step 3 into the equation:
\[ q\left(\sqrt{\frac{2qV}{m}}\right)B = \frac{m\left(\frac{2qV}{m}\right)}{r} \]
Simplifying this gives:
\[ qB\sqrt{\frac{2qV}{m}} = \frac{2qV}{r} \]

6. **Solving for Radius \(r\):**
Rearranging to find \(r\):
\[ r = \frac{2qV}{qB\sqrt{\frac{2qV}{m}}} \]
Simplifying further:
\[ r = \frac{2\sqrt{qV}}{B\sqrt{2q/m}} \]

7. **Finding \(\sin \alpha\):**
The angle of deviation \(\alpha\) relates to the width of the magnetic field \(d\) and the radius \(r\) of the circular path:
\[ \sin \alpha = \frac{d}{r} \]
Substituting for \(r\):
\[ \sin \alpha = \frac{dB\sqrt{2q/m}}{2\sqrt{qV}} \]

### Final Expression:
Thus, the value of \(\sin \alpha\) is:
\[ \sin \alpha = \frac{dB}{2\sqrt{qV/m}} \]