***PLEASE ANSWER 7,8,9*** Any charged particle with nonzero spin has a magnetic

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***PLEASE ANSWER 7,8,9***

Any charged particle with nonzero spin has a magnetic moment and acts like a little bar magnet. This is the source of the unfortunate name "spin", left over from when physicists were still trying to impress classical behavior on quantum particles. Classically, a spinning ball of electric charge would have a circular current and therefore a magnetic moment given by sigma LA summed over all circular currents in the particle. Spin is a form of angular momentum and changes by plusminus h. For electrons, protons, neutrons, spin = plusminus h/2. For shorthand, particles with half-integer spin are called fermions. mu _n = -1.913 mu _N where mu _N = e_ft/2m_ is the nuclear magneton. mu _p = 2.793 mu _N The proton radius is about R_p = 0.88 fm. (1 fm, or Fermi, is one femtometer, 10^-15 m. If we let the total charge of the proton be located at the surface, the charge density would be sigma = e/4 pi R^2 _y. [1] For a proton spinning with angular velocity omega, a ring of radius r and width R_y d theta would have a charge of q = sigma (m omega ^2 R_p d0) [2] and a current of i = sigma (m omega ^3 R_p d theta) [3] Rewrite r in terms of R_p, and theta. sin theta = r/R_p or sin theta R_p = r For smaller angle, sin theta = theta So, r = R_p theta [4] The magnetic moment for a current loop with area A is given by mu = iA. Write the expression for mu for your current loop using your current above and the area in terms of the expression for r in [3]. Replace r with the expression in [3] and factor all constants to the front of the expression. mu - i A = sigma (m omega ^3 R_p d theta) (pi (R_p theta ^2)) = sigma (pi omega ^3 R^3 _p theta ^2 d theta) [5] To add up the magnetic moments for the entire surface, add up magnetic moments for all the rings by integrating over theta from 0 to pi. mu ' = integral ^pi _0 sigma (pi m omega ^3 R^3 _p theta ^2 d theta) = [sigma (pi m omega ^3 R^3 _p theta ^3)/3]^pi _0 = [sigma (pi m (pi omega R_o)^3)/3] [6] Experimentally, mu _p = 1.410607 J/T. Set that value equal to the expression above and solve for omega. w = 0.428773/mR^2 _p sigma [7] Calculate the velocity at the equator of the proton using the radius of the proton and the omega from [6]. w = 0.428773/m (0.88 * 10^-15 m) ^3 * sigma ? ? ? [8] Comment on the possibility of the result in [7]. How would the velocity result change if the charge were distributed over the entire volume of the proton instead of just the surface?. [9] What does this mean for the validity of the spinning particle model for accounting for the particle's magnetic moment?

Explanation / Answer

Ans 7. Centrepetal force=m*v^2/r; Now since this force is magnetic, hence the force applied is lorenzian i.e. F=qvB.

Thus equating the 2 equations we get, Bqv=m*v^2/r;

v=Brq/m where B=The magnetic field, r=given radius, q=1.6*10^-19C, m=mass of proton=1836*9.1*10^-31kG.

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