Question

Derive Eq. 8.39. [Hint: Treat the lower loop as a magnetic dipole.]

Short answer

The equation 8.39 is derived as$F_{mag}=\frac{3\pi }{2}{\mu }_0{l}_a{l}_b\frac{a^2{b}^{2}h}{{(b^2+h^2)}^{{\displaystyle \frac{3}{2}}}}$ .

Step-by-step solution

Expression for the magnetic field of the loop:

Write the expression for the magnetic field of the loop.

$\mathit{B}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}{\mathbf{l}}_{\mathbf{b}}}{\mathbf{2}}\frac{{\mathbf{b}}^{\mathbf{2}}}{{\mathbf{(}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{h}}^{\mathbf{2}}\mathbf{)}}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}\mathit{z}$

Here,${\mathit{l}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{b}}$are the carrying currents.

Derive equation 8.39 as :3π2μ0IaIba2b2h(b2+h2)32dz=mag

Write the expression for the magnetic force experienced by the loop.

$$\begin{gathered} F_{mag}=\mu (\nabla B) \\ =\mu \left(\frac{dB}{dB}\right).............\left(1\right) \end{gathered}$$

Here, $\mu$is the magnetic dipole moment of the loop, which is given as:

$$\begin{gathered} \mu =l_{a}A \\ =l_a\pi a^2 \end{gathered}$$

Substitute the known values in equation (1).

$$\begin{gathered} F_{mag}=\left(l_a\pi a^2\right)\frac{d}{dh}\left[\left(\frac{{\mu }_0{l}_b}{2}\right)\left(\frac{\begin{array}{c}\\ b^2\end{array}}{{(b^2+h^2)}^{{\displaystyle \frac{3}{2}}}}\right)\right] \\ =\left[\left(l_a\pi a^2\right)\left(\frac{{\mu }_0{l}_b}{2}\right)\right]\frac{d}{dh}\left(\frac{\begin{array}{c}\\ b^2\end{array}}{{(b^2+h^2)}^{{\displaystyle \frac{3}{2}}}}\right) \end{gathered}$$

$$\begin{gathered} =\frac{{\mu }_0\pi {\alpha }^2{l}_a{l}_b}{2}\left[\left(\frac{-3}{2}\right){(b^2+h^2)}^{\frac{-5}{2}}\left(2h\right)b^2-0\right] \\ =\frac{{\mu }_0\pi {\alpha }^2{l}_a{l}_b}{2}\left(\frac{3h{b}^2}{{(b^2+h^2)}^{{\displaystyle \frac{5}{2}}}}\right) \end{gathered}$$

The gravitational force acting on the loop is$F=m_{a}g$ .

On further solving,

$$\begin{gathered} {\mathit{F}}_{\mathbf{m}\mathbf{a}\mathbf{g}}\mathbf{=}\frac{{\mathbf{\mu }}_{\mathbf{0}}\mathbf{\pi }{\mathbf{a}}^{\mathbf{2}}{\mathbf{l}}_{\mathbf{a}}{\mathbf{l}}_{\mathbf{b}}}{\mathbf{2}}\mathbf{}\left(-\frac{3h{b}^2}{{(b^2+h^2)}^{{\displaystyle \frac{5}{2}}}}\right) \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{}\mathbf{}\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}{\mathit{\mu }}_{\mathbf{0}}{\mathit{l}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{b}}\frac{{\mathbf{a}}^{\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}\mathbf{h}}{{\mathbf{(}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{h}}^{\mathbf{2}}\mathbf{)}}^{{\displaystyle \frac{\mathbf{5}}{\mathbf{2}}}}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}{\mathit{m}}_{\mathbf{a}}\mathit{g} \end{gathered}$$

Therefore, equation 8.39 is derived as$\frac{\mathbf{3}\mathbf{\pi }}{\mathbf{2}}{\mathit{\mu }}_{\mathbf{0}}{\mathit{l}}_{\mathbf{a}}{\mathit{l}}_{\mathbf{b}}\frac{{\mathbf{a}}^{\mathbf{2}}{\mathbf{b}}^{\mathbf{2}}\mathbf{h}}{{\mathbf{(}{\mathbf{b}}^{\mathbf{2}}\mathbf{+}{\mathbf{h}}^{\mathbf{2}}\mathbf{)}}^{{\displaystyle \frac{\mathbf{3}}{\mathbf{2}}}}}\mathit{d}\mathit{z}\mathbf{=}{\mathit{m}}_{\mathbf{a}}g$ .