= Solution
Write the total <velocity> as $\mathbf U=\Omega y\widehat{\mathbf x}+\mathbf u$ and the <magnetic field> as $B_0\widehat{\mathbf x}+\mathbf b$. For solenoidal fields the <resistive induction equation> is $(\partial_t+\mathbf U\cdot\nabla)\mathbf B=\mathbf B\cdot\nabla\mathbf U+\eta\nabla^2\mathbf B$. The <first-order smoothing approximation> drops the fluctuating terms $\mathbf b\cdot\nabla\mathbf u-\mathbf u\cdot\nabla\mathbf b$, retaining shear exactly. The uniform field is not stretched by the background shear because $\partial_x(\Omega y)=0$. The other retained stretching terms are $\mathbf b\cdot\nabla(\Omega y\widehat{\mathbf x})=\Omega b_y\widehat{\mathbf x}$ and $B_0\partial_x\mathbf u$. Hence
$$
\partial_t\mathbf b+\Omega y\partial_x\mathbf b=\Omega b_y\widehat{\mathbf x}+B_0\partial_x\mathbf u+\eta\nabla^2\mathbf b.
$$
The same phase cancellation as in part (i), together with $\nabla^2 e^{i\mathbf k\cdot\mathbf x}=-K^2e^{i\mathbf k\cdot\mathbf x}$, proves that a response can be taken as $\operatorname{Re}(\mathbf c(t)e^{i\mathbf k(t)\cdot\mathbf x})$, where
$$
\boxed{\dot{\mathbf c}=\Omega c_y\widehat{\mathbf x}+ik_xB_0\mathbf v-\eta K^2\mathbf c.}
$$
If $\mathbf k\cdot\mathbf c=0$ initially, its derivative is $-\eta K^2\mathbf k\cdot\mathbf c$, so magnetic solenoidality is preserved.
For $j=y,z$ the response equation is $\dot c_j+\eta K^2c_j=ik_xB_0v_j$. Its integrating-factor solution is
$$
c_j(t)=e^{-\eta\int_{t_0}^tK^2ds}c_j(t_0)+ik_xB_0\int_{t_0}^t e^{-\eta\int_\tau^tK^2ds}v_j(\tau)d\tau.
$$
The <velocity> amplitudes and <wavevector> vary on the shear timescale. In the fast-diffusion limit $\eta(k_x^2+k_y^2)\gg|\Omega|$, and after the free transient, the <integral> is concentrated within time $(\eta K^2)^{-1}$ of $t$. Replacing the slowly varying factors there by their current values gives the <quasistatic magnetic response of a helical shearing wave>:
$$
\boxed{c_y\simeq\frac{ik_xB_0v_y}{\eta K^2},\qquad c_z\simeq\frac{ik_xB_0v_z}{\eta K^2}.}
$$
<Complex conjugation> introduces a minus sign in $c_j^*$. With precisely the EMF normalization specified in the question,
$$
\mathcal E=\operatorname{Re}\left[-\frac{ik_xB_0}{\eta K^2}(v_yv_z^*-v_zv_y^*)\right]=\boxed{-\frac{B_0H(t)}{\eta K^2}\ \text{to leading order}.}
$$
A literal spatial average of the product of the two real Fourier waves would be half this quantity; no extra factor of $1/2$ belongs in the question's defined $\mathcal E$.
Use $A=k_x^2+k_z^2$ and part (i) to obtain $\mathcal E(t)\simeq-B_0H_0 A/[\eta(A+\Omega^2k_x^2t^2)^2]$. For the time <integral>, take the driven response without a free magnetic transient and suppose the fast-diffusion approximation holds through the contributing interval, in particular at $t=0$; $\eta k_x^2\gg|\Omega|$ is a sufficient uniform condition. For $\Omega k_x\ne0$, substitute $z=|\Omega k_x|t/\sqrt A$. Since $\int_{-\infty}^{\infty}(1+z^2)^{-2}dz=\pi/2$, this gives
$$
\boxed{\int_{-\infty}^{\infty}\mathcal E(t)dt\simeq-\frac{\pi B_0H_0}{2\eta|\Omega k_x|\sqrt{k_x^2+k_z^2}}.}
$$
The <integral> identity follows immediately from $z=\tan\chi$, reducing it to $\int_{-\pi/2}^{\pi/2}\cos^2\chi\,d\chi$. If $k_x=0$, the forcing and $H$ vanish and the driven EMF is zero. If $\Omega=0$ but $H_0\ne0$, the leading EMF is constant rather than time-integrable; the finite <integral> relies on nonzero shear.
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