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$$e=\frac{d\Phi}{dt}=\frac{\pi r^2\mu_0N}{l}\frac{dl}{dt}$$

The equation above is the solution to the problem. The $r$ refers to the solenoids radius, $l$ is the solenoids length and $N$ is the amount of loops around the solenoid.

Why doesnt the distance between the loop and the solenoid affect the electromagnetic field (EMF) induced in the loop? Does this mean induced EMF is always equally strong no matter the distance from the source?

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The formula you are referring to is only an approximation for very long solenoids. The magnetic field outside but close to the very long solenoid is very weak, so that only the magnetic field inside the solenoid contributes significantly to the EMF in the loop in this limit case: $$\oint_{\partial A} \vec E\cdot d\vec s=\int_A (\nabla \times \vec E)\cdot d\vec A=-\frac{\partial}{\partial t}\int_A \vec B\cdot d\vec A\approx-\frac{\partial}{\partial t}\int_{A_{Solenoid}} \vec B\cdot d\vec A$$ which has indeed an amplitude independent of the loop radius, given for example a sinusoidal excitation of the solenoid.

The situation changes, if you widen the loop considerably. Then you will also capture part of the magnetic flux that returns outside the solenoid. If you finally let the loop tend to infinity, then the total magnetic flux through its cross section will tend to zero, and with it the induced EMF.

See this reference for a calculation of the solenoid field. Although this is still a short solenoid, it illustrates the basic observation that the magnetic flux density drops abruptly when crossing the boundary between inside and outside the solenoid. For the long solenoid it is going to be even more pronounced. By the way, the software used is FEMM, a free software I have also often used to understand electro-/magnetostatic problems. You could build a long solenoid yourself with it by only a small effort. However, the tool is limited to 2D problems, but the solenoid is one if you choose cylindrical coordinates.

On german Wikipedia there is also this graph, which even gives a better impression about the flux density jump when crossing the windings.

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  • $\begingroup$ Ok so if I understand it correctly you're saying that with infinitely large loops the total emf will approach 0 since all the returning flux will be captured by the loop, and this flux's direction is opposite to the flux inside the solenoid? $\endgroup$ – Joel __ Apr 1 at 18:35
  • $\begingroup$ @Joel__: Correct. $\endgroup$ – oliver Apr 2 at 8:41
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The $\vec B$-field outside a very long solenoid is very nearly $0$, except very close to the coil, as discussed here:

Templin, J.D., 1995. Exact solution to the field equations in the case of an ideal, infinite solenoid. American Journal of Physics, 63(10), pp.916-920.

Thus, the amount of flux captured by the outside loop is very nearly independent of its radius because, i.e. in the extra area between $r_2$ and $r_1$ outside the solenoid, $\vec B\approx 0$, so not net flux is added when your increase $r$.

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