General Magnetics (06/05/2020)

Consider the circle to be a general distance $x$ from the wire. A general chord parallel to the wire is parameterised in polar coordinates and a point of intersection of the circle and chord is:

$x_c = x + a \cos{\theta}$ $y_c = a \sin{\theta}$

The area of a strip along the chord of thickness $\delta x_c$ is:

$\delta A = 2y_c \delta x_c$ $\implies \delta A = -2 a^2 \sin^2{\theta} \delta{\theta}$

The flux through this elementary strip is:

$\delta \Phi = \frac{\mu_o I}{2 \pi \left(x+ a \cos{\theta}\right)}\delta A$

$\implies \delta \Phi = -\frac{2 a^2 \mu_o I \sin^2{\theta} \delta{\theta}}{2 \pi \left( x+a \cos{\theta}\right)}$

Considering only the magnitude, the negative sign can be dropped:

$\implies \delta \Phi = \frac{a^2 \mu_o I}{ \pi}\left(\frac{\sin^2{\theta} \ \delta{\theta}}{ x+a \cos{\theta}}\right)$

Taking the limit as $\theta \to 0$ leads to:

$d \Phi = \frac{a^2 \mu_o I}{ \pi}\left(\frac{\sin^2{\theta} \ d\theta}{ x+a \cos{\theta}}\right)$

The magnitude of the total flux through the circle is:

$\Phi = \frac{a^2 \mu_o I}{ \pi} \int_{0}^{\pi} \frac{\sin^2{\theta} \ d\theta}{ x+a \cos{\theta}} = \mu_o I x \ \dots (1)$

Keeping in mind $x >> a$

$\Phi = \frac{a^2 \mu_o I}{ \pi x} \int_{0}^{\pi} \frac{\sin^2{\theta} \ d\theta}{ 1+\frac{a \cos{\theta}}{x}}$

$\implies \Phi = \frac{a^2 \mu_o I}{ \pi x} \int_{0}^{\pi} \frac{\sin^2{\theta} \ d\theta}{ 1+\frac{a \cos{\theta}}{x}} \approx \frac{a^2 \mu_o I}{ \pi x} \int_{0}^{\pi} \sin^2{\theta} \ d\theta$

$\Phi \approx \frac{a^2 \mu_o I}{2 x}$

The rest of the solution proceeds according to that of @Steven Chase . It is noteworthy that if the assumption that $x >> a$ is not considered, then the final result is independent of $x$ .

Magnetic flux density:

$B = \frac{\mu_0 I}{2 \pi x}$

Flux linkage:

$\lambda = \pi a^2 B = \frac{\mu_0 I a^2}{2 x}$

Induced voltage $V$ :

$V = \frac{d \lambda}{dt} = \frac{d \lambda}{dx} \frac{dx}{dt} = -\frac{\mu_0 I a^2}{2 x^2} \dot{x}$

Power dissipated:

$P = \frac{V^2}{r} = \Big( \frac{\mu_0 I a^2}{2 x^2} \Big)^2 \, \frac{\dot{x}^2}{r}$

Rearranging for $\dot{x}$ :

$\dot{x} = \frac{2 x^2}{\mu_0 I a^2} (P r)^{1/2}$