It is sensitive to E-fields, but since it is shielded the only reasonable place for the E-Fields to interact with the conductor is at the very top and the wavelength must be of the same order as the size of the gap for it to couple in any energy.
That is presumably several orders of magnitude difference in wavelengths and your circuit that is analyzing the output will be looking at the larger (presumably) return signal form the dominant magnetic coupling. The E-Field that is of the same wavelength as the H-Field will not "see" that gap.
What confuses me is that the flux derivative is inside the surface
integral.
The partial derivative inside the integral comes from Leibniz Integral Rule (detailed below).
Consider generalized form of Maxwell-Faraday Equation:
$$\oint E \cdot d\ell = -\int_\Sigma \frac{\partial B}{\partial t} \cdot dA$$
This is true for any path \$\partial \Sigma\$, which is any closed-contour bounds the surface \$\Sigma\$.
Now remember generalized form of Leibniz Integral Rule:
$$\frac{d}{dt}\int^{a(x)}_{b(x)} f(x, t) \ dt = f(x, b(x)) \ \frac{d}{dt}b(x) - f(x, a(x)) \
\frac{d}{dt} a(x) + \int^{a(x)}_{b(x)} \frac{\partial}{\partial t}f(x, t) \ dt$$
Did you notice that the bounds of the integral above are not constants?
The right-side term of the generalized Maxwell-Faraday Equation is a surface integral (and, of course, integral around \$\partial \Sigma\$ is a line integral) and the partial derivative inside this integral indicates that any \$\partial \Sigma\$ path is time-dependent. That's why we write Maxwell-Faraday Equation in generalized form because we cannot guarantee that any \$\partial \Sigma\$ is constant.
Now let's look at Leibniz Rule again when the bounds are constants. The first two term of the right-side becomes zero and the integral takes its own special form:
$$\frac{d}{dt}\int^{a}_{b} f(x, t) \ dt = \int^{a}_{b} \frac{\partial}{\partial t}f(x, t) \ dt$$
From this, we can make a conclusion: If the path \$\partial \Sigma\$, which bounds the surface \$\Sigma\$, does not change over time, Maxwell-Faraday Equation turns into:
$$\oint E \cdot d\ell = -\frac{d}{dt}\int_\Sigma B \cdot dA$$
NOTE: I wrote Leibniz Integral Rule for single dimension and made explanations over it just to make things simpler, but the same thing applies for higher dimensions.
Best Answer
It is technically true that very intense magnetic fields can do awful things to a human brain, but what you're asking about is current flow. In your example there is no return path for current flow, the circuit is not complete, and you are safe. If there was a second lead that could complete the circuit, then current could flow.
Of course, I should point out that if your example creates an extreme voltage, the path could possibly complete itself, through the air, in the form of an arc. Also, don't go thinking you can just put on thick rubber shoes and grab a high voltage line - the rubber forms a kind of capacitance, through which a lethal AC current can still flow to ground.