Help with Magnetic Interactions

[jughead4466]

Particles having mass=m and charge = Q travel parallel to the z axis, forming a beam of radius = R and uniform charge density = p. To keep the beam focused, an external uniform magnetic field, B, parallel to the z axis is provided, and the beam is made to rotate with a constant, uniform angular velocity = w.

4. Use Gauss' Law to find the radial electric field in the beam on a cylinder of radius r<R

6a. Find the tangential velocity of a particle in the beam at r<R

8b. Find the total (electric and magnetic) force on a particle at r<R

 

[Nate810]

The radial electric field in the beam on a cylinder of radius r<R can be found using Gauss' Law: E_r = \frac{pQ}{2\pi r}.The tangential velocity of a particle in the beam at r<R is given by v_t = rw, where w is the angular velocity of the beam.The total force on a particle at r<R is equal to the vector sum of the electric and magnetic forces. The electric force is given by F_e = \frac{pQ^2}{2\pi r^2}, and the magnetic force is given by F_m = \frac{Qv_tB}{c}, where c is the speed of light. Thus, the total force is given by F = \frac{pQ^2}{2\pi r^2} + \frac{QrwB}{c}.

 

[ogb p]

I would be happy to assist you with your inquiries about magnetic interactions. Firstly, let's break down the given information to better understand the situation. We have a beam of particles with mass m and charge Q traveling parallel to the z axis. The beam has a radius of R and a uniform charge density of p. To keep the beam focused, we have an external uniform magnetic field B parallel to the z axis, and the beam is rotating with a constant, uniform angular velocity w.

To find the radial electric field in the beam on a cylinder of radius r<R, we can use Gauss' Law. This law states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space (ε0). In this case, we can consider a cylindrical surface with radius r and length L, enclosing a portion of the beam. The charge enclosed by this surface would be given by q = pπr^2L, where p is the charge density and πr^2L is the volume of the enclosed cylinder. Therefore, the electric flux through this surface would be given by ΦE = q/ε0 = (pπr^2L)/ε0. Since the electric flux is also equal to the integral of the electric field over the surface, we can equate these two equations to find the radial electric field at any point r<R as E = (pπr)/ε0.

To find the tangential velocity of a particle in the beam at r<R, we can use the equation for centripetal force, Fc = mw^2r, where m is the mass of the particle and w is the angular velocity. This force is balanced by the magnetic force, Fm = QvB, where v is the tangential velocity of the particle and B is the magnetic field. Setting these two equal to each other and solving for v, we get v = (mw)/Q.

Finally, to find the total force on a particle at r<R, we need to consider both the electric and magnetic forces. Since the electric force is given by Fe = qE = (pπr^2L)E, and the magnetic force is given by Fm = QvB = (Qv)B, we can add these two forces together to get the total force on the particle as F = (pπr^2L)