# Tag Info

## Hot answers tagged electromagnetic-induction

Accepted

### Why does a changing magnetic flux induce an EMF in "opposite direction"?

Besides the answer "because that's how the maths works", there is a conservation of energy argument. Consider a dipole magnet, which you drop through a metal tube. As it falls, it creates a ...
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### Why does a changing magnetic flux induce an EMF in "opposite direction"?

Actually the Lenz law is connected to energy conservation (one of the most fundamental "law" of nature). So let us suppose you have a fixed conducting loop and a movable magnet. If you move ...
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### Can you use electrical appliances inside a Faraday's Cage?

Can you use electrical appliances inside a Faraday's Cage? Yes. I have read that if a charge was to be placed inside the cavity of a Faraday's Cage, its electric field would be canceled by the field ...
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### Magnetic force is a no work force. Why is it doing work here?

This answer adds little – except, perhaps, some linking passages – to the comments already made. The rod's velocity is shared by the free electrons in it, and therefore these experience a magnetic ...
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### Can you use electrical appliances inside a Faraday's Cage?

However, using this logic, I find it hard to understand why scientists often conduct experiments in rooms designed as Faraday Cages with electrical equipment, in order to reduce interference from ...
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Accepted

### Am I misunderstanding the concept of magnetic flux?

$$\phi_{B}=\iint \vec{B} \cdot \vec{da}$$ If your surface is changing, as with the moving rod example, the bounds of the integral need to vary with time Assuming that the B field function exists in ...
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### Doubt regarding the calculation of EMF due to self-inductance and mutual inductance

As suggested by Jan, your sentence about self-induction The self-induced EMF is equal to the negative of the change in the net magnetic flux passing through the inductor is wrong. A valid sentence ...
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### Force components on dipole falling through copper ring

Your error is assuming that $$\int_0^{2\pi} \left[ IRB_z \hat{\rho}\right] \; d\phi = \left[\int_0^{2\pi} IRB_z\; d\phi \right]\hat{\rho}.$$ This is because at different points on the wire, the ...
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### Why is the flux linkage of one coil with respect to the other always equal if they are coaxial

No, it doesn't need to assume the fields outside the solenoid are zero. The use of long solenoids is convenient because you can compute the flux easily and do the integrals easily. But you could also ...
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