= Solution
In the Lorentz law, $q$ is signed electric charge, $\mathbf E$ and $\mathbf B$ are the electric and <magnetic fields> at the bead, and the <velocity> in $q\mathbf V\times\mathbf B$ is its full laboratory <velocity>. Here $\mathbf V=v\hat{\mathbf r}+w\hat{\boldsymbol\theta}$. Since $\hat r\times\hat z=-\hat\theta$ and $\hat\theta\times\hat z=\hat r$,
$$
\boxed{F_{B,r}=qBw,\qquad F_{B,\theta}=-qBv,\qquad F_{B,z}=0.}
$$
The tangential component integrated around an instantaneous <circle> of nearly fixed radius gives
$$
\boxed{W_\theta=\int_0^{2\pi}F_{B,\theta}R\,d\theta=-2\pi RqvB.}
$$
This is the stated tangential <circulation> <work> expression in the $w\gg v$ limit. \b[It is tangential <circulation> <work>.] The <tangential circulation work versus particle work> distinction matters because the bead actually follows an expanding spiral. The radial <displacement> may be small, but its <force> is proportional to $w$, so its <power> is of the same order as the tangential <power>. In fact
$$
P_{B,r}=qBwv,\qquad P_{B,\theta}=-qBvw,\qquad\boxed{\mathbf F_B\cdot\mathbf V=0.}
$$
The cancellation expresses the fact that <magnetic forces do no work>. Ignoring radial <work> while retaining the first-order tangential effect would incorrectly conclude that a <magnetic field> supplies net particle <energy>.
The <electric field> giving the same tangential <force> would be $\boxed{\mathbf E_{\rm eq}=-vB\hat\theta}$. This is a force-equivalent field for the moving guide, not evidence that a real circulating <electric field> has appeared in the laboratory merely because the loop expands in a static <magnetic field>. An <electromotive force> is the tangential driving-force <circulation> per unit charge, with units of <energy> per charge. For a material circuit moving with local <velocity> $\mathbf v_{\rm loop}$,
$$
\mathcal E=\oint(\mathbf E+\mathbf v_{\rm loop}\times\mathbf B)\cdot d\mathbf l.
$$
Orient the ring counterclockwise with positive normal $+\hat z$. Its flux is $\Phi=B\pi R^2$, so $\dot\Phi=2\pi BRv$. The tangential <circulation> therefore gives
$$
\boxed{\mathcal E=W_\theta/q=-2\pi BRv=-\frac{d\Phi}{dt}.}
$$
It is the motional term that matters here. For a fixed loop, the <circulation> of the actual <electric field> instead follows the partial time <derivative> of the <magnetic field>. The distinction between moving-wire magnetic <forces> and induction in a stationary loop is essential to the flux rule.<The Feynman Lectures on Physics, Volume II by Richard Feynman and others>[II.17–1, equations (17.1)–(17.3) and the distinction between moving circuits and changing fields; II.17–2, limits on applying the flux rule]{ref}
For the <magnetic power balance on an expanding circular guide>, the guide must supply an additional inward reaction $-qBw\hat r$ opposing the radial magnetic <force>. With radial guide <speed> $v$, this contribution to its <power> on the bead is \b[$P_{\rm guide}^{(B)}=-qBvw$], equal to the tangential magnetic <power>. The radial magnetic <power> is positive by the same amount, so the total magnetic <power> still vanishes; the mechanical exchange is associated with the <moving mechanical constraint>.
If “the ring's <work>” is meant as its full <work> on a finite-mass bead, the source statement needs a further qualification. For a frictionless bead of <mass> $m_b$ with constant imposed radial <speed>, its radial acceleration is $-w^2/R$, so the complete radial guide reaction and <power> are
$$
N_r=-\frac{m_bw^2}{R}-qBw,\qquad P_{\rm guide}=N_rv=-\frac{m_bvw^2}{R}-qBvw.
$$
The first term is present even when $B=0$ and describes the familiar change of tangential <kinetic energy> on an expanding guide. The requested $-qBvw$ is exactly the additional magnetic-field-dependent term, not that full reaction <power>. No bead <mass> or extra agency fixing $w$ is supplied, so it would be unjustified to assign a numerical total mechanical <power> or silently discard its centripetal contribution.
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