Diagaram benda bebas sistem:
Gambar 2
Seperti yang terlihat pada gambar di atas, besar tegangan kedua tali berbeda. Mari kita mencari arah torsi akibat tegangan tali dan momen dipole $\displaystyle ( \mu )$, terhadap titik $\displaystyle O$.
Tinjau tali kiri,
$$
\begin{aligned}
\overrightarrow{\tau _{A}} & =T_{A}\cos \theta \hat{k} \times R(\hat{\jmath })\\
\overrightarrow{\tau _{A}} & =-T_{A} R\cos \theta \hat{\imath }\ \ \ \dots (1)
\end{aligned}$$
Tinjau tali kanan,
$$
\begin{aligned}
\overrightarrow{\tau _{B}} & =T_{B}\cos \theta \hat{k} \times R( -\hat{\jmath })\\
\overrightarrow{\tau _{B}} & =T_{B} R\cos \theta \hat{\imath }\ \ \ \dots (2)
\end{aligned}$$
Tinjau torsi akibat momen dipole,
$$
\begin{aligned}
\overrightarrow{\tau _{\mu }} & =\vec{\mu } \times \vec{B}\\
\overrightarrow{\tau _{\mu }} & =NAI\hat{k} \times B_{0}( -\hat{\jmath })\\
\overrightarrow{\tau _{\mu }} & =NAB_{0} I\hat{\imath }\ \ \ \dots (3)
\end{aligned}$$
Dari (1), (2), dan (3) resultan torsi yang bekerja adalah
$$
\begin{aligned}
\Sigma \vec{\tau } & =0\\
\overrightarrow{\tau _{A}} & =\overrightarrow{\tau _{B}} +\overrightarrow{\tau _{\mu }}\\
T_{A} R\cos \theta \hat{\imath } & =T_{B} R\cos \theta \hat{\imath } +NAB_{0} I\hat{\imath }\\
T_{A} R\cos \theta -T_{B} R\cos \theta & =NAB_{0} I\\
T_{A} -T_{B} & =\frac{\pi R^{2} B_{0} I}{R\cos \theta }\\
T_{A} -T_{B} & =\frac{\pi B_{0} IR}{\cos \theta } \ \ \ \dots (4)
\end{aligned}$$
Selanjutnya, mari kita tinjau resultan gaya dalam arah sumbu $\displaystyle y$,
$$
\begin{aligned}
\Sigma F_{y} & =0\\
T_{A}\cos \theta +T_{B}\cos \theta & =Mg\\
( T_{A} +T_{B})\cos \theta & =Mg\\
T_{A} +T_{B} & =\frac{Mg}{\cos \theta }\ \ \ \dots (5)
\end{aligned}$$
Jumlahkan (4) dan (5),
$$
\begin{aligned}
2T_{A} & =\frac{\pi B_{0} IR}{\cos \theta } +\frac{Mg}{\cos \theta }\\
T_{A} & =\pi B_{0} IR+Mg\ \ \ \dots (\text{Jawaban})
\end{aligned}$$
Kurangkan (4) dan (5),
$$
\begin{aligned}
-2T_{B} & =\frac{\pi B_{0} IR-Mg}{\cos \theta }\\
T_{B} & =Mg-\pi B_{0} IR\ \ \ \dots (\text{Jawaban})
\end{aligned}$$
Gambar 1