a charge particle q enters a uniform magnetic filed B directed outwards. It is deflected by a distance d after travelling a horizontal distance a.The magnitude of momentum of particle is

When a charged particle moves through a magnetic field, it experiences a force that causes it to change direction. This phenomenon is governed by the Lorentz force law, which states that the force acting on a charged particle is proportional to the charge, the velocity of the particle, and the magnetic field strength. Let's break down the situation you've described to find the magnitude of the momentum of the particle.

Understanding the Motion in a Magnetic Field

When the charged particle \( q \) enters the magnetic field \( B \), it experiences a magnetic force given by:

F = q(v × B)

Here, \( v \) is the velocity of the particle, and \( B \) is the magnetic field vector. The direction of the force is perpendicular to both the velocity and the magnetic field, causing the particle to move in a circular path. The radius of this circular motion can be derived from the balance of the magnetic force and the centripetal force required to keep the particle in circular motion.

Finding the Radius of Curvature

The centripetal force required to keep the particle moving in a circle is given by:

F_c = \frac{mv^2}{r}

Setting the magnetic force equal to the centripetal force, we have:

qvB = \frac{mv^2}{r}

From this equation, we can solve for the radius \( r \):

r = \frac{mv}{qB}

Analyzing the Deflection

As the particle travels a horizontal distance \( a \) and is deflected by a distance \( d \), we can relate these quantities to the radius of curvature. The angle \( \theta \) through which the particle is deflected can be approximated for small angles as:

\(\theta \approx \frac{d}{r}\)

Since the particle travels a horizontal distance \( a \), we can express the relationship between \( a \), \( r \), and \( \theta \) as:

a = r \theta

Substituting the expression for \( \theta \), we get:

a = r \frac{d}{r} = d

This indicates that the distance \( d \) is directly related to the radius of curvature and the horizontal distance \( a \). Now, we can express \( r \) in terms of \( a \) and \( d \):

r = \frac{a}{d}

Calculating the Momentum

Now that we have the radius of curvature, we can substitute this back into the equation for the radius:

\(\frac{a}{d} = \frac{mv}{qB}\)

Rearranging this gives us:

mv = \frac{a q B}{d}

The quantity \( mv \) represents the momentum \( p \) of the particle. Therefore, we can express the momentum as:

p = \frac{a q B}{d}

Final Result

In summary, the magnitude of the momentum of the charged particle \( q \) after traveling a horizontal distance \( a \) and being deflected by a distance \( d \) in a uniform magnetic field \( B \) is given by:

p = \frac{a q B}{d}

This formula encapsulates the relationship between the distance traveled, the charge of the particle, the strength of the magnetic field, and the deflection experienced by the particle. Understanding these relationships is crucial in fields such as electromagnetism and particle physics.