5. Force on Moving Ring in Radial Field
Step 1: Magnetic Flux Definition
The problem describes a “radial magnetic field” originating from the origin, falling off such that the result contains $(x^2+a^2)$. This geometry corresponds to the field of a point magnetic source (monopole-like flux) where $B \propto 1/r^2$.
The solid angle subtended by the ring at position $x$ is $\Omega = 2\pi (1 – \cos\theta) = 2\pi (1 – \frac{x}{\sqrt{x^2+a^2}})$.
Flux $\Phi = \beta \Omega = 2\pi \beta (1 – \frac{x}{\sqrt{x^2+a^2}})$.
Step 2: Induced EMF
As the ring moves with velocity $v = dx/dt$, the flux changes:
$$ \mathcal{E} = -\frac{d\Phi}{dt} = -2\pi \beta \frac{d}{dx}\left( 1 – \frac{x}{(x^2+a^2)^{1/2}} \right) v $$Using the derivative quotient rule:
$$ \frac{d}{dx} \left( \frac{-x}{\sqrt{x^2+a^2}} \right) = – \frac{\sqrt{x^2+a^2} – x \frac{x}{\sqrt{x^2+a^2}}}{x^2+a^2} = – \frac{a^2}{(x^2+a^2)^{3/2}} $$ $$ \mathcal{E} = 2\pi \beta v \frac{a^2}{(x^2+a^2)^{3/2}} $$Step 3: Power and Force
The current in the ring is $I = \mathcal{E}/R$. Power dissipated is $P = I^2 R = \mathcal{E}^2/R$.
The mechanical power required to maintain velocity $v$ is $F v$.
$$ F v = \frac{\mathcal{E}^2}{R} \implies F = \frac{\mathcal{E}^2}{v R} $$ $$ F = \frac{1}{v R} \left[ \frac{2\pi \beta a^2 v}{(x^2+a^2)^{3/2}} \right]^2 $$
$$ F = \frac{4\pi^2 a^4 \beta^2 v}{R (x^2+a^2)^3} $$