Question 8.6: A small current loop L1 with magnetic moment 5 az A · m^2 is...

The torque T_{2} on the loop L_{2} is due to the field B_{1} produced by loop L_{1}. Hence

T_{2}=m_{2}\times B_{1}

Since m_{1} for loop L_{1} is along a_{z}, we find B_{1} using eq. (8.22):

B=\frac{\mu_{o}m}{4\pi r^{3}}(2\cos\theta a_{r}+\sin\theta a_{\theta })

B_{1}=\frac{\mu_{o}m_{1}}{4\pi r^{3}}(2\cos\theta a_{r}+\sin\theta a_{\theta })

Using eq. (2.23):

T=m\times B=Q_{m}\ell\times B

we transform m_{2} from Cartesian to spherical coordinates:

m_{2}=3a_{y}=3(\sin\theta \sin\phi a_{r}+\cos\theta \sin\phi a_{\theta} +\cos\phi a_{\phi })

At (4,-3,10)

r=\sqrt{4^{2}+(-3)^{2}+10^{2}}=5\sqrt{5}

\tan\theta =\frac{\rho }{z}=\frac{5}{10}=\frac{1}{2}\rightarrow \sin\theta=\frac{1}{\sqrt{5}},       \cos\theta=\frac{2}{\sqrt{5}}

\tan\phi =\frac{y}{x}=\frac{-3}{4}\rightarrow \sin\phi =\frac{-3}{5},       \cos\phi =\frac{4}{5}

Hence

B_{1}=\frac{4\pi\times 10^{-7}\times5}{4\pi625\sqrt{5}}\left(\frac{4}{\sqrt{5}}a_{r}+\frac{1}{\sqrt{5}}a_{\theta}\right)=\frac{10^{-7}}{625}(4a_{r}+a_{\theta})

m_{2}=3\left[-\frac{3a_{r}}{5\sqrt{5}}-\frac{6a_{\theta}}{5\sqrt{5}}+\frac{4a_{\phi}}{5}\right]

and

T=\frac{10^{-7}(3)}{625(5\sqrt{5})}(-3a_{r}-6a_{\theta}+4\sqrt{5}a_{\phi})\times(4a_{r}+a_{\phi})

=4.293\times10^{-11}(-8.944a_{r}+35.777a_{\theta}+21a_{\phi})

=-0.384a_{r}+1.536a_{\theta}+0.9015a_{\phi}nN\cdot m