Deriving Particle Momentum in Circular Magnetic Field

This question asks us to find the expression for the momentum of a charged particle moving in a circular path within a uniform magnetic field. The magnetic field is perpendicular to the plane of the circle.

Understanding the Forces Involved

When a charged particle with charge '$Q$' moves with velocity '$v$' in a magnetic field '$B$', it experiences a magnetic force (Lorentz force) given by the formula:

$$ \vec{F} = Q(\vec{v} \times \vec{B}) $$

In this specific case, the velocity vector '$\vec{v}$' is perpendicular to the magnetic field vector '$\vec{B}$'. Therefore, the magnitude of the magnetic force simplifies to:

$$ F = QvB $$

This magnetic force is responsible for keeping the particle moving in a circle of radius '$R$'. This means the magnetic force is acting as the centripetal force required for circular motion. The formula for centripetal force is:

$$ F_c = \frac{mv^2}{R} $$

where '$m$' is the mass of the particle and '$v$' is its speed.

Equating Forces for Circular Motion

Since the magnetic force provides the centripetal force, we can set the two expressions equal:

$$ F = F_c $$

$$ QvB = \frac{mv^2}{R} $$

Calculating the Momentum

The momentum '$p$' of a particle is defined as the product of its mass and velocity:

$$ p = mv $$

We can rearrange the force balance equation ($QvB = \frac{mv^2}{R}$) to find an expression for momentum ($mv$).

First, let's simplify the equation by canceling one '$v$' from both sides (assuming $v \ne 0$):

$$ QB = \frac{mv}{R} $$

Now, we can solve for '$mv$' by multiplying both sides by '$R$':

$$ QBR = mv $$

Since momentum $p = mv$, we have:

$$ p = QBR $$

Conclusion

The momentum of the particle moving in a circle of radius '$R$' with speed '$v$' in a uniform magnetic field '$B$' (perpendicular to the plane of motion) is $QBR$.

Summary Table