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Given a circular wire loop of radius r, resistance R enclosing a magnetic field perpendicular to the plane of the loop that increases with time ($B=\alpha t$), I have calculated the induced current $I=\frac{\pi r^2 \alpha}{R}$. Now, consider an ideal voltmeter connecting between points P and Q on the loop, which have an angle $90^{\circ}$ between them. The question I am struggling with is what the voltmeter reads.

My attempt so far is to note that $V(b) - V(a) = \int_a^b \nabla V \cdot d\vec{l} = -\int_a^b \vec{E} \cdot d\vec{l}$ and I would like to plug in the electric field, which I tried to get using $\nabla \times E = -\alpha$ but of course I do not know $vec{E}$ explicitly.

In addition to asking how to solve this problem. My question about physics is: how is the potential gradient distributed when the current is uniform?

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  • $\begingroup$ What the voltmeter reads depends 100 percent on how you wire the leads of the voltmeter. Induction leads to a non-conservative field, so the total potential difference depends on the path around an area of the field. What the path is depends on where you run the lead wires of the voltmeter trough the field volume. The answer can be anything from minus a very large number to plus a very large number (up to the point where materials start breaking down in the electric field). $\endgroup$ – CuriousOne May 1 '15 at 0:45
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The voltage between a and b is$IR=πr^2α/4.$ this is trivial it is a quarter of the Voltage. The Electric field is not generated from the $-∇V$ but from $-dA/dt $

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  • $\begingroup$ That is what I thought. But the solution in the book (Griffiths) says the answer is $\alpha r^2 / 2$ $\endgroup$ – user44816 May 1 '15 at 12:41
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Just like the curiousone told you, since the curl of E is not zero, you're dealing with a non-conservative field. So if you consider the potential difference to be the work done by E to get from the point P to Q it does matter the way you take. Griffiths answer is based on the trajectory to go from P to Q in a straight line.

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