# Magnetic flux

by ElmorshedyDr
Tags: flux, magnetic
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 HW Helper Thanks PF Gold P: 4,855 Then you use the Blv law. The Blv law applies even in situations where there is moving media and Maxwell's equation del x E = - dB/dt does not apply. In your case the area is formed by the motion of the wire: delta area = lv per second, so dA/dt = B dA/dt = Blv.
P: 403
 Quote by phy_infinite Also if the magnetic field doesn't change, a changing area also changes the flux.

I can't understand how changing area changes flux since the geometrical shape of wire is constant so the flux should be constant!!
P: 75
 Quote by phy_infinite If the magnetic field is constant, then as you move the wire, the flux is constant and does not change.
Here, I should have said the flux changes, since the area changes.
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PF Gold
P: 4,855
 Quote by ElmorshedyDr I can't understand how changing area changes flux since the geometrical shape of wire is constant so the flux should be constant!!
Flux is area times B. If area changes, flux changes and there is an emf generated.

Area changes as vl. Think of the wire sweeping the area behind it as it moves in the perpendicular B field. dA/dt = lv.
P: 75
 Quote by ElmorshedyDr I can't understand how changing area changes flux since the geometrical shape of wire is constant so the flux should be constant!!
It's more intuitive and accurate really to think about the Lorentz force law. For a wire of length l moving through a constant magnetic field at some velocity, it can be found using the Lorentz force law that EMF = Blv.
 P: 853 for a physical picture imagine a simple dc electric motor, you have stationary magnets and a rotor which has wires through it that form coils, these wires are constantly moving and " cutting" through the field of those magnets and current is being produced.
P: 75
 Quote by ElmorshedyDr I can't understand how changing area changes flux since the geometrical shape of wire is constant so the flux should be constant!!
To see this more rigorously, imagine you're pulling on a wire of length l to the right side of the computer screen and the wire itself is pointing up on the computer screen in a magnetic field that is pointing into the screen. Then the emf is $$\varepsilon = \oint \mathbf{f} \cdot d \mathbf{l}$$ Where $\mathbf{f}$ is the force per unit charge. The magnetic force will add a vertical component to the electrons in the wire so they will be moving at an angle towards the top right corner of the screen. The electrons would then be moving upward with speed u and to the right with speed v. Now, the magnetic force will also have a component to the left side of the screen. This force to the left would be equal to QuB. You would then have to pull with force per unit charge uB to the right. The total distance travelled by a charge is $\frac{l}{cos \theta}$ Therefore the work done per unit charge would be $$\int \mathbf{f}_{pull} \cdot d \mathbf{l} = (uB) (\frac{l}{cos \theta}) sin \theta = vBl = \varepsilon$$

Fortunately, this gives a result equivalent to the flux rule for motional emf.

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