Question

Hi, so I’m stuck on a question where I have to prove that the radius of the curvature of a particle is equal to (1/BQ) * (2*Ek*m)^0.5, using the centripetal force equation, kinetic energy and F=BQv to do so. Can someone help me please?

Answer 1

Given:

1. Centripetal force equation. The magnitude of the centripetal force (the force that makes a body follow a curved path) on an object of mass m moving at tangential speed v along a path with radius of curvature r is:

$$F=\frac{mv^2}{r}.$$

2. Kinetic energy. The kinetic energy of a point object (an object so small that its mass can be assumed to exist at one point) is:

$$E_k=\frac{mv^2}{2}.$$

3. Consider the case of a particle of charge Q entering a uniform magnetic field B with its initial velocity vector of magnitude v perpendicular to the field. The constant magnitude of the magnetic force is given by

$$F=BQv.$$

This force can’t change the magnitude of particle velocity but can deflect the particle continuously along a curvilinear path. The tangential component of the force will be zero as force and velocity are mutually perpendicular. Therefore, the normal component will be equal to F itself. This will always act perpendicular to velocity at each point. Therefore, F is the magnitude of the centripetal force.

Proof:

1. From given 1 and 3:

$$\frac{mv^2}{r}=BQv,$$ $$r=\frac{mv}{BQ}.$$

2. From given 2:

$$v=\sqrt{\frac{2E_k}{m}}.$$

3. From proof 1 and 2:

$$r=\frac{m}{BQ}\sqrt{\frac{2E_k}{m}}=\frac{\sqrt{2E_km}}{BQ}.$$