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If a rod of length $L$ is moving at uniform velocity, $v$, perpendicular to the magnetic field, the EMF induced in the rod is $\mathcal E=BLv$.

I was wondering if the same thing would apply for a horizontal and vertical solenoid moving at uniform velocity perpendicular to the direction of magnetic field.

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2 Answers 2

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One should ask, where does the equation come from? it is always helpful for these kind of questions to understand the reasoning behind coming up for that equation, thus being able to apply it to different applications.

What we know (what Faraday taught), is that changing the "magnetic field" induces an electric field, mathematically written as

$$e.m.f = -\frac{\Delta\Phi}{\Delta t}$$

with $\Phi$ being the element representing how much magnetic "field" passes through an area ($\Phi = BA$, where $B$ is the magnetic field and $A$ is the area thorugh which the magnetic field passes), this is called flux. This equation basically means that changing the amount of magnetic field that goes thorugh an area should change for a voltage difference to be induced. This can be done either by:

  1. changing the flux

  2. changing the area

The expression $e = Blv$ used in academic text books is to do with a situation like below:

enter image description here credits

If a rod moves only by itself, it is true that the charges inside will feel a force and be opposite charges will be shifted to 2 different ends of the rod (not really, just the electrons moving but it has the same effect), but the induced emf is not as simple as $Blv$.

So, about the solenoids, you would not have any induced voltage if it's just moving through the field with constant speed, at $t=0$, the opposite charges will gather at opposite ends, but then everything stops, just like that!

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I'm not sure what a "horizontal and vertical solenoid" is, but I shall assume that its cross-section is rectangular with sides of the rectangle lying in the ±x and ±y alignments, and that the magnetic field is in the $z$ direction (that is parallel to the length of the solenoid).

Suppose that the solenoid is moved steadily in the $x$ direction. Emfs will be induced in the $–y$ direction, so as we go cyclically around the rectangle the emfs ($Blv$) in the two $±y$ aligned sides will be in opposite senses (one will be clockwise and the other, anticlockwise as seen from the same end of the solenoid) and will cancel. There will be no emfs in the $±x$ aligned sides. This will be the case for all turns of the solenoid, so there will be no overall emf.

You can reach the same conclusion by noting that there is no change in the flux linking the solenoid if it is moved as described.

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