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nice explanation, Philip. I hadn't really thought about this before. So the idea is that the loop is rigid, and is not being rotated, so the velocity at all parts of the loop is the same, and the magnetic field is uniform. And the equation for the emf due to the magnetic field is:
[tex]\oint (\vec{v} \wedge \vec{B}) \ \cdot d \vec{L} [/tex]
But since the integrand is a constant vector, then we can simply write it as the gradient of some scalar field, for example if:
[tex]\vec{v} \wedge \vec{B} = q \hat{z} [/tex]
(where q is just a number, not charge or anything) then we can write it as the gradient of the scalar field [itex]\phi=qz[/itex] And since we can write it in this form, the emf is:
[tex]\oint \nabla (\phi) \ \cdot d \vec{L} [/tex]
But luckily for us, this is the definition of:
[tex]\oint d \phi [/tex]
And this must equal zero because it is around a closed loop. Therefore the emf in this situation must equal zero. (Sorry if this is using too much vector calculus). I am interested in the explanation using cutting of flux and flux linkage. Because I never really got comfortable with those concepts.
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