The neutron is a particle with zero charge. Nevertheless, it has a nonzero magne

Question

The neutron is a particle with zero charge. Nevertheless, it has a nonzero magnetic moment with a z?component of 9.66 x 10-27 Am2 . This can be explained by the internal structure of the neutron. A substantial body of evidence indicates that a neutron is composed of three fundamental particles called quarks: an “up” (u) quark of charge +2e/3, and two “down” (d) quarks of charge ?e/3. The combination of the three quarks produces a net charge of zero. If the quarks are in motion, they can produce a nonzero magnetic moment. As a very simple model, assume the u quark moves in a counterclockwise circular path and the d quarks move in a clockwise circular path, all of radius r and all with the same speed v.

a. What is the current due to the circulation of the u quark?  

b. What is the magnitude of the magnetic moment due to the circulating u quark?

c. What is the magnitude of the magnetic moment of the three?quark system.

d. With what speed v must the quarks move if this model is to reproduce the magnetic moment of the neutron? Use r = 1.20 x 10-15m (the radius of the neutron) for the radius of the orbit.

Explanation / Answer

a.) The perimeter of the circle is given by 2r

if a particle moves around with velocity v m/s it is going to cross a point on the circle (say, the topmost point of the figure) v / 2r times in a second. If the particle holds a charge q, the net effective charge crossing this point per second (which is nothing but the current) is nothing but qv/2r

therefore I =qv/2r

for u quark, charge = + 2e/3 , therefore current because of it is ev/3r where, r is given as 1.2 x 10-15 m

b.) Magnetic moment due to u quark = Current due to u quark x area = (ev/3r ) x r2 = evr /3 A.m2

c.) Current due to one d quark = (- e/3)(-v) /2r = ev / 6r

(since, charge is -e/3 and velocity is negative since it is circulating in the opposite direction)

Current due to both d quarks = 2 x (ev/6r) = ev /3r

Note that the current is positive, which means it is in the direction of the current because of u quark. So, the currents add up to give bigger current and hence magnetic moment.

Net current due to 3 quark system = ev/ 3r + ev /3r = 2ev/ 3r

Net magnitude of three quark system is (2ev/ 3r ) x r2 = 2evr / 3 A.m2

d.) 2evr /3 = 9.66 x 10-27

v = 3 x 9.66 x 10-27 /2er

v = 1.5 x 9.66 x 10-27 / ( 1.60217662 × 10-19 x 1.2 x 10-15 )

v = 7.536622 x 107   m/s

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