Quark model of the neutron. The neutron is a particle with zero net charge. Neve

Question

Quark model of the neutron. The neutron is a particle with zero net charge. Nevertheless, it has a nonzero magnetic dipole moment of magnitude mu = 9.66 times 10^-27 A middot m^2. This can be explained by assuming the neutron consists of three quarks: an 'up' (u) quark u with charge +2e/3, and two 'down' (d) quarks, each with charge -e/3. In a simple toy model, the quarks all move with fixed speed upsilon in a circular orbit of radius r = 1.20 times 10^-15 m, but in different directions, as shown. The figure lies in the xy plane with the z-direction pointing out of the paper. a) Derive an expression for the magnetic dipole moment of the system in terms of the given variables (magnitude and direction). (b) With what speed upsilon must the quarks move to reproduce the dipole moment of the neutron? (c) The neutron is now placed in a magnetic field B = 2.0 times 10^-3 T x. What is the torque vector experienced by the dipole? Which way does the dipole move, if at all?

Explanation / Answer

magnetic dipole moment is mu = i*A

current due to up quark is during one complete revolution i = q/t = (2e/3) / (2*pi*r/v) = 2ev/(6*pi*r) = ev/(3*pi*r)

current due to each down quark is i = (e/3) / (2*pi*r/v) = ev/(6*pi*r)

due to two down quarks is 2*ev/(6*pi*r) = ev/(3*pi*r)

so net current is Inet = (ev/(3*pi*r)) + (ev/(3*pi*r)) = 2*ev/(3*pi*r)

So magnetic moment is mu = Inet*A = (2*ev/(3*pi*r))*pi*r^2 = 2*evr/3 = (2/3)*e*v*r

direction is along + Z-axis

so mu = (2/3)*e*v*r along +Z-axis

b) mu = 2*e*v*r/3

9.66*10^-27 = 2*1.6*10^-19*v*1.2*10^-15/3

v = 7.54*10^7 m/sec

c) Torque T = mu*B*sin(theta) = 9.66*10^-27*2*10^-3*sin(90)

T = 1.932*10^-29 N-m along +Y-axis

direction of torque or axis of rotation is about +y-axis

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