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enter image description hereFor a straight wire moving perpendicular to a uniform magnetic field in velocity v, $$ε=\dfrac{\Delta \Phi}{\Delta t}=\dfrac{BLv \Delta t}{\Delta t}$$ So the induced e.m.f in the conductor as it moves through the magnetic field is: $ε = BLv$. However, isn't the rate of magnetic flux cut by the straight wire itself zero? The straight wire has the same amount of flux cutting through it every second. The area where the wire cut across the magnetic flux is changing with time, but it's not a closed area. Does Faraday's law apply to unclosed areas too?

Also, Faraday's law states that 'the magnitude of the emf induced in a circuit is proportional to the rate of change with time t of the magnetic flux $\Phi$ that cuts across the circuit', but where is the circuit here?

Sancol.
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Maxine
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1 Answers1

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I note that this is a question you have modified as the previous one was closed due to lack of clarity.

There are numerous variations of related questions on this website with What is the difference between induced emf and motional emf? as an example which advises you to take the motional emf route.

The straight wire has the same amount of flux cutting through it every second.
It is not the magnetic flux through the physical area of the wire which is being used rather it is the flux through the red area in your diagram.
The top, right-side, bottom of the red area, and the wire on the left hand define the loop, which does not have to be conducting, through which the magnetic flux is calculated.
So it is the area which is changing as the wire moves to the right.

You might find the MIT publication Faraday's Law of Induction of interest, particularly 10.2 Motional EMF page 10.7, 10.9.2 Loop Changing Area page 10.18, and 10.9.3 Sliding Rod page 10.19.

Farcher
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