91Ó°ÊÓ

1. The vertical axis scale is set by $U_s=2.0×{10}^{-4}J$.
2. The counterclockwise rotation from $\mathrm{Ï•}=0$
3. The rotational kinetic energy is$K_0=6.7×{10}^{-4}J$.

By using Eq.28-39 and information from the given Fig.28-52, we can find the orientation energyU of dipole at$\mathrm{Ï•}=0$. By using this value of in the equation of themechanical energy, we can find the value of the orientation energy U_turnof dipole at the turning point. By using this value of U_turn in the formula for the angle $\mathrm{Ï•}$, we can find themaximum value of $\mathrm{Ï•}$.

Formula:

From Eq. 28-39, the orientation energy of dipole is

$U=\mathrm{μ}⃗.B⃗$

The mechanical energy is given by

$K+U=K_0+(-\mathrm{μ}B)$

The angle is given by

role="math" localid="1662906003335" $\mathrm{ϕ}=-co{s}^{-1}\left(\frac{U_{turn}}{\mathrm{μ}B}\right)$

From Eq. 28-39, the orientation energy of dipole is

$\begin{array}{l}U=\stackrel{→}{\mathrm{μ}B}\\ U=-\mathrm{μ}B\cos \mathrm{ϕ}\end{array}$

So, $\mathrm{Ï•}=0$corresponds to the lowest point in the Fig.28-51. The mechanical energy is given by

$K+U=K_0+(-\mathrm{μ}B)$

Substitute all the value in the above equation.

$K+U=6.7×{10}^{-4}J+(-5.0×{10}^{-4}J)$

Where, from the given Fig.28-52, $U=-5.0×{10}^{-4}J.$

$K+U=1.7×{10}^{-4}J$

The turning point occurs where K=0, which gives

$U_{turn}=1.7×{10}^{-4}J$

So, the corresponding angle is given by

$$\begin{gathered} \mathrm{ϕ}=-{\cos }^{-1}\left(\frac{1.7×{10}^{-4}J}{5.0×{10}^{-4}J}\right) \\ \mathrm{ϕ}=-{70}^{∘} \end{gathered}$$

Since the angle is negative, it means it is measured in clockwise direction with respect to the negative horizontal axis. It implies that the angle is $\mathrm{ϕ}={110}^{∘}$with respect to the positive horizontal axis in the diagram.