A straight conductor slides with constant velocity v

A straight conductor slides with constant velocity $v$ across a circular conducting ring of radius $R$, in a uniform magnetic field of induction $B$ perpendicular to the plane of the ring. Find the electromotive force in the circuit when the conductor is at distance $d$ from the centre.

Step 1: The enclosed area

Take the origin at the centre of the ring, with the $x$ axis along the direction of motion. A chord at distance $x$ from the centre cuts off a circular segment of area

\[ S(x) = R^{2}\arccos\frac{x}{R}-x\sqrt{R^{2}-x^{2}}. \tag{1} \]

Step 2: The flux

With $\vec B$ perpendicular to the plane and antiparallel to the outward normal, $\Phi(x) = -B\,S(x)$.

Step 3: Differentiate the segment area

The two terms of (1) contribute

\[ \frac{d}{dx}\left(R^{2}\arccos\frac{x}{R}\right) = -\frac{R^{2}}{\sqrt{R^{2}-x^{2}}}, \]

\[ \frac{d}{dx}\left(x\sqrt{R^{2}-x^{2}}\right) = \sqrt{R^{2}-x^{2}}-\frac{x^{2}}{\sqrt{R^{2}-x^{2}}} = \frac{R^{2}-2x^{2}}{\sqrt{R^{2}-x^{2}}}, \]

so that, subtracting, \[\frac{dS}{dx} = \frac{-R^{2}-\bigl(R^{2}-2x^{2}\bigr)}{\sqrt{R^{2}-x^{2}}} = \frac{-2\bigl(R^{2}-x^{2}\bigr)}{\sqrt{R^{2}-x^{2}}} = -2\sqrt{R^{2}-x^{2}}. \tag{2}\]

Step 4: Apply Faraday's law

Since $\Phi = -BS$, equation (2) gives $\dfrac{d\Phi}{dx} = -B\dfrac{dS}{dx} = 2B\sqrt{R^{2}-x^{2}}$, and with $\dfrac{dx}{dt} = v$: $\mathcal{E} = -\frac{d\Phi}{dt} = -\frac{d\Phi}{dx}\cdot\frac{dx}{dt}$.

\[ \mathcal{E} = -2Bv\sqrt{R^{2}-x^{2}}. \]

Step 5: The answer is the motional EMF

The chord at distance $d$ has length $\ell = 2\sqrt{R^{2}-d^{2}}$, so the result is exactly

\[ \bigl|\mathcal{E}\bigr| = B\,v\,\ell, \]

the standard formula for a straight conductor of length $\ell$ moving with speed $v$ perpendicular to a field $B$. Only the part of the conductor actually inside the ring contributes, which is why the answer shrinks as $d$ grows.

Step 6: Verification

\[d = 0 \text{ (the conductor on a diameter): } \bigl|\mathcal{E}\bigr| = 2BvR, \text{ the maximum} \;\;\checkmark\]

\[d\to R:\ \bigl|\mathcal{E}\bigr|\to0, \text{ the chord shrinking to a point} \;\;\checkmark\]

\[\text{Sign check: } \frac{dS}{dx}\lt 0 \text{ — the enclosed segment shrinks as } x \text{ increases} \;\;\checkmark\]