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The neutron is a particle with zero charge. Nonetheless, it has a nonzero magnetic moment with z-component $\mathbf{9}\mathbf{.}\mathbf{6}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{27}}\mathbf{A}\mathbf{.}{\mathbf{m}}^{\mathbf{2}}$. 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 鈥渦p鈥� (u) quark, of charge +2e/3 , and two 鈥渄own鈥� (d) quarks, each of charge . The combination of the three quarks produces a net charge zero. If the quarks are in motion, they can produce a nonzero magnetic moment. As a very simple model, suppose the quark moves in a counterclockwise circular path and the quarks move in a clockwise circular path, all of radius r and all with the same speed v in figure. (a) Determine the current due to the circulation of the u quark. (b) Determine the magnitude of the magnetic moment due to the circulating u quark. (c) Determine the magnitude of the magnetic moment of the three quark system. (Be careful to use the correct magnetic moment directions.) (d) With what speed v must the quarks move if this model is to reproduce the magnetic moment of the neutron? Use r = $\mathbf{1}\mathbf{.}\mathbf{20}\mathbf{脳}{\mathbf{10}}^{\mathbf{-}\mathbf{15}}\mathbf{m}$(the radius of the neutron) for the radius of the orbits.

Step by step solution

01

Definition of magnetic field

The term magnetic field may be defined as the area around the magnet behave like a magnet.

02

Determine the various term for quarks

The current due to the circulation of the quarks is

$\begin{array}{r}{\mathrm{I}}_{\mathrm{v}}=\frac{\mathrm{dq}}{\mathrm{dt}}\\ {\mathrm{I}}_{\mathrm{v}}=\frac{\mathrm{螖辩}}{\mathrm{螖迟}}\\ \mathrm{螖迟}=\frac{2\mathrm{蟺谤}}{\mathrm{v}}\\ {\mathrm{I}}_{\mathrm{v}}=\frac{{\mathrm{q}}_{\mathrm{u}}\mathrm{V}}{2\mathrm{蟺谤}}\\ \mathrm{Or}\\ {\mathrm{I}}_{\mathrm{u}}=\frac{\mathrm{ev}}{3\mathrm{蟺谤}}\end{array}$

Hence, the current due to the circulation of the u quark is${\mathrm{I}}_{\mathrm{v}}=\frac{\mathrm{ev}}{3蟺\mathrm{r}}$

The magnitude of the magnetic moment can be calculated as

$\begin{array}{r}{\mathrm{渭}}_{\mathrm{u}}={\mathrm{l}}_{\mathrm{u}}\mathrm{A}\\ =\frac{\mathrm{ev}}{3\mathrm{蟺谤}}{\mathrm{蟺谤}}^2\\ =\frac{\mathrm{evr}}{3}\end{array}$

Hence, the magnitude of the magnetic moment due to the circulating u quark is .

${\mathrm{渭}}_{\mathrm{v}}=\frac{\mathrm{evr}}{3}$
The magnitude of the magnetic moment of the three quark system

$\begin{array}{r}{\mathrm{渭}}_{\text{total}}={\mathrm{渭}}_{\mathrm{u}}+{\mathrm{渭}}_{\mathrm{d}}\\ {\mathrm{渭}}_{\text{total}}=\frac{2\mathrm{evr}}{3}\end{array}$

Hence, the magnitude of the magnetic moment of the three quark system is ${\mathrm{渭}}_{\text{total}}=\frac{2\mathrm{evr}}{3}$.

The speed can be calculated by the relation

$\begin{array}{r}\mathrm{v}=\frac{3\mathrm{渭}}{2\mathrm{er}}\\ \mathrm{v}=\frac{3\left(9.66脳{10}^{鈭�27}\mathrm{A}鈰�{\mathrm{m}}^2\right)}{2\left(1.60脳{10}^{鈭�19}\mathrm{C}\right)\left(1.20脳{10}^{鈭�15}\mathrm{m}\right)}\\ \mathrm{v}=7.55脳{10}^7\mathrm{m}/\mathrm{s}\end{array}$

Hence, the speed is$7.55脳{10}^7\mathrm{m}/\mathrm{s}$

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