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I don't understand how a rod translating or rotating perpendicular to a magnetic field could produce an emf. To produce an emf, we need a changing magnetic flux. But here, there is no change in magnetic flux, area or the angle between normal vector and magnetic field. Then why would there be a potential difference generated?

An aeroplane with a wing span of $50 \ \text{m}$ is flying horizontally over a place where the vertical component of the Earth's magnetic field is $2 \times 10^{−4} \ \text{Wb/m}^2$. The potential difference between the tips of the wings is $1 \ \text{V}$. What is the speed of the aeroplane?

I have the same issue with this question.

In the case of the translating rod, I understand that the Lorentz force can push the electrons to one side. But then, why is it a case of Faraday's Law?

I am aware that similar questions have been asked here, but none of them have cleared my doubts. It is the rotating rod that confuses me the most. Why does the rod "remember" the area it swept?

M. A.
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Neerada Nas
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1 Answers1

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I don't claim that what follows in quotes should be called 'Faraday's law', but it seems to cover most cases, including awkward ones like the rotating disc (with magnetic field normal to the plane of the disc and with brushes touching axle and rim)...

"When a conductor cuts magnetic flux or the flux through a circuit changes, an emf is induced in the conductor or circuit. Its magnitude is equal to the rate of cutting, or to the rate of change, of flux."

Rate of cutting is calculated from the motional emf: $$d\mathscr E=(\vec v \times \vec B).d \vec {\mathscr l}.$$ In the case of the rotating disc one integrates along a radius and then notes that all the radii are electrically in parallel.

Philip Wood
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