Chapter 8: Q11P (page 378)

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

Short Answer

Expert verified

The equation 8.39 is derived as $F_{m a g} = \frac{3 π}{2} μ_{0} l_{a} l_{b} \frac{a^{2} b^{2} h}{{(b^{2} + h^{2})}^{\frac{3}{2}}}$ .

Step by step solution

Expression for the magnetic field of the loop:

Write the expression for the magnetic field of the loop.

$B = \frac{μ_{0} l_{b}}{2} \frac{b^{2}}{{(b^{2} + h^{2})}^{\frac{3}{2}}} z$

Here, $l_{a} l_{b}$ 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.

$F_{m a g} = μ (\nabla B) = μ (\frac{d B}{d B}) . . . . . . . . . . . . . (1)$

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

$μ = l_{a} A = l_{a} π a^{2}$

Substitute the known values in equation (1).

$F_{m a g} = (l_{a} π a^{2}) \frac{d}{d h} [(\frac{μ_{0} l_{b}}{2}) (\frac{\begin{matrix} b^{2} \end{matrix}}{{(b^{2} + h^{2})}^{\frac{3}{2}}})] = [(l_{a} π a^{2}) (\frac{μ_{0} l_{b}}{2})] \frac{d}{d h} (\frac{\begin{matrix} b^{2} \end{matrix}}{{(b^{2} + h^{2})}^{\frac{3}{2}}})$

$= \frac{μ_{0} π α^{2} l_{a} l_{b}}{2} [(\frac{- 3}{2}) {(b^{2} + h^{2})}^{\frac{- 5}{2}} (2 h) b^{2} - 0] = \frac{μ_{0} π α^{2} l_{a} l_{b}}{2} (\frac{3 h b^{2}}{{(b^{2} + h^{2})}^{\frac{5}{2}}})$

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

On further solving,

$F_{m a g} = \frac{μ_{0} π a^{2} l_{a} l_{b}}{2} (- \frac{3 h b^{2}}{{(b^{2} + h^{2})}^{\frac{5}{2}}}) = \frac{3 π}{2} μ_{0} l_{a} l_{b} \frac{a^{2} b^{2} h}{{(b^{2} + h^{2})}^{\frac{5}{2}}} = m_{a} g$

Therefore, equation 8.39 is derived as $\frac{3 π}{2} μ_{0} l_{a} l_{b} \frac{a^{2} b^{2} h}{{(b^{2} + h^{2})}^{\frac{3}{2}}} d z = m_{a} g$ .

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

Short Answer

Step by step solution

Expression for the magnetic field of the loop:

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

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