Figure 28-52 gives the orientation energy Uof a magnetic dipole in an external magnetic field $\overrightarrow{\mathbf{B}}$, as a function of angle $\varphi$ between the directions $\overrightarrow{B}$, of and the dipole moment. The vertical axis scale is set by $U_s=2.0\times {10}^{-4}J$. The dipole can be rotated about an axle with negligible friction in order that to change $\varphi$. Counterclockwise rotation from $\varphi =0$yields positive values of $\varphi$, and clockwise rotations yield negative values. The dipole is to be released at angle $\varphi =0$with a rotational kinetic energy of $6.7\times {10}^{-4}J$, so that it rotates counterclockwise. To what maximum value of $\mathit{\varphi }$will it rotate? (What valueis the turning point in the potential well of Fig 28-52?)

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

By using Eq.28-39 and information from the given Fig.28-52, we can find the orientation energyU of dipole at$\varphi =0$. 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 $\varphi$, we can find themaximum value of $\varphi$.

Formula:

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

$U=\mu ⃗.B⃗$

The mechanical energy is given by

$K+U=K_0+(-\mu B)$

The angle is given by

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

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

$\begin{array}{l}U=\overrightarrow{\mu B}\\ U=-\mu B\cos \varphi \end{array}$

So, $\varphi =0$corresponds to the lowest point in the Fig.28-51. The mechanical energy is given by

$K+U=K_0+(-\mu B)$

Substitute all the value in the above equation.

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

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

$K+U=1.7\times {10}^{-4}J$

The turning point occurs where K=0, which gives

$U_{turn}=1.7\times {10}^{-4}J$

So, the corresponding angle is given by

$$\begin{gathered} \varphi =-{\cos }^{-1}\left(\frac{1.7\times {10}^{-4}J}{5.0\times {10}^{-4}J}\right) \\ \varphi =-{70}^{\circ } \end{gathered}$$

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 $\varphi ={110}^{\circ }$with respect to the positive horizontal axis in the diagram.