A particle accelerated by a potential difference `V` flies through a uniform transverse magnetic field with induction B. The field occupies a region of space d in thickness. Prove that the angle a through which the particle deviates from the initial direction of its motion is given by.
`alpha=sin^(-1)(dBsqrt((q)/(2 Vm)))`
where `m` is the mass of the particle.

Options

A. `theta=sin^(-1)(dBsqrt((qV)/(2m)))`
B. `theta=sin^(-1)(dBsqrt((q)/(2mV)))`
C. `theta=tan^(-1)(dBsqrt((qV)/(2m)))`
D. `theta=tan^(-1)(dBsqrt((q)/(2mV)))`

Answer

To prove the angle \( \alpha \) through which the particle deviates from its initial direction of motion, we will follow these steps:

### Step 1: Understand the motion of the particle
A charged particle is accelerated through a potential difference \( V \). The kinetic energy gained by the particle can be expressed as:
\[ KE = qV \]
where \( q \) is the charge of the particle.

### Step 2: Relate kinetic energy to momentum
The kinetic energy can also be expressed in terms of momentum \( p \):
\[ KE = \frac{p^2}{2m} \]
where \( m \) is the mass of the particle. Equating the two expressions for kinetic energy gives:
\[ qV = \frac{p^2}{2m} \]

### Step 3: Solve for momentum
From the equation above, we can solve for momentum \( p \):
\[ p = \sqrt{2mqV} \]

### Step 4: Determine the radius of the circular path
When the particle enters the magnetic field, it moves in a circular path due to the magnetic force acting as the centripetal force. The magnetic force is given by:
\[ F = qvB \]
where \( v \) is the velocity of the particle and \( B \) is the magnetic field strength. The centripetal force required for circular motion is:
\[ F = \frac{mv^2}{r} \]
Setting these two forces equal gives:
\[ qvB = \frac{mv^2}{r} \]
From this, we can solve for the radius \( r \):
\[ r = \frac{mv}{qB} \]

### Step 5: Relate the angle of deviation to the geometry of the motion
The particle travels a distance \( d \) in the magnetic field. The angle \( \alpha \) can be related to the radius \( r \) and the thickness \( d \) of the magnetic field region. From the geometry of the circular motion, we have:
\[ \sin(\theta) = \frac{d}{r} \]
where \( \theta \) is the angle of deviation. Therefore, we can express \( \alpha \) as:
\[ \alpha = \sin^{-1}\left(\frac{d}{r}\right) \]

### Step 6: Substitute for \( r \)
Substituting the expression for \( r \) into the equation for \( \alpha \):
\[ \alpha = \sin^{-1}\left(\frac{d}{\frac{mv}{qB}}\right) \]

### Step 7: Substitute for \( v \)
We know that \( v \) can be expressed in terms of kinetic energy:
\[ v = \sqrt{\frac{2qV}{m}} \]
Substituting this into the equation for \( \alpha \):
\[ \alpha = \sin^{-1}\left(\frac{d \cdot qB}{m \cdot \sqrt{\frac{2qV}{m}}}\right) \]

### Step 8: Simplify the expression
This simplifies to:
\[ \alpha = \sin^{-1}\left(\frac{dB\sqrt{q}}{\sqrt{2Vm}}\right) \]

### Final Result
Thus, we have shown that:
\[ \alpha = \sin^{-1}\left(dB\sqrt{\frac{q}{2Vm}}\right) \]
This completes the proof.

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