Calculation of Change in Magnetic Flux Linkage of a moving wire

[Jy158654]

Homework Statement

A straight wire of length 0.20m moves at a steady speed of 3.0m/s at right angles to a magnetic field of flux density 0.10T. Use Faraday's law to determine the e.m.f. induced across the ends of a wire.

Relevant Equations

E= Nd Φ/dt but N=1 so E= dΦ/dt

Can anyone explain the above answer to me? What does the "change in area" mean?

 

[etotheipi]

Usually when this argument is presented it uses a diagram like this, to make it easier to visualise:

Faraday tells you that $\mathcal{E} = - \frac{d\Phi}{dt}$, where $\Phi$ is the magnetic flux through a surface which is bounded by a closed curve around which the EMF is $\mathcal{E}$. In this case, you can consider the EMF around a curve running through the rectangle including the moving wire, and the magnetic flux through the two dimensional surface that the curve encloses.

Then you use that, in this case, $\Phi = \vec{B} \cdot \vec{A} = BA \implies \frac{d\Phi}{dt} = B \frac{dA}{dt} = BLv$ to deduce that $\mathcal{E} = - BLv$ (I take that the normal to the surface, $\hat{n}$, points in the same direction as the magnetic field).

Another way you can derive the same thing is to consider an equilibrium of the magnetic (due to external $\vec{B}$ field) and electric forces (due to charge separation inside the wire) on a charge element inside the wire. You will have$$q\vec{v} \times \vec{B} + q\vec{E} = 0 \implies \vec{E} = -\vec{v} \times \vec{B}$$If $\vec{v}$ and $\vec{B}$ are orthogonal, then the result is $\vec{E} = -vB \hat{x}$, where we've just oriented our Cartesian axes so that $\hat{x}$ is parallel to the wire. Then$$\mathcal{E} = \int \vec{E} \cdot d\vec{r} = \int_0^L E_x dx = -BLv$$

 

[Jy158654]

Usually when this argument is presented it uses a diagram like this, to make it easier to visualise:

View attachment 267929

Faraday tells you that $\mathcal{E} = - \frac{d\Phi}{dt}$, where $\Phi$ is the magnetic flux through a surface which is bounded by a closed curve around which the EMF is $\mathcal{E}$. In this case, you can consider the EMF around a curve running through the rectangle including the moving wire, and the magnetic flux through the two dimensional surface that the curve encloses.

Then you use that, in this case, $\Phi = \vec{B} \cdot \vec{A} = BA \implies \frac{d\Phi}{dt} = B \frac{dA}{dt} = BLv$ to deduce that $\mathcal{E} = - BLv$ (I take that the normal to the surface, $\hat{n}$, points in the same direction as the magnetic field).

Another way you can derive the same thing is to consider an equilibrium of the magnetic (due to external $\vec{B}$ field) and electric forces (due to charge separation inside the wire) on a charge element inside the wire. You will have$$q\vec{v} \times \vec{B} + q\vec{E} = 0 \implies \vec{E} = -\vec{v} \times \vec{B}$$If $\vec{v}$ and $\vec{B}$ are orthogonal, then the result is $\vec{E} = -vB \hat{x}$, where we've just oriented our Cartesian axes so that $\hat{x}$ is parallel to the wire. Then$$\mathcal{E} = \int \vec{E} \cdot d\vec{r} = \int_0^L E_x dx = -BLv$$

Hello. Thank you for your reply! I understand the formula you derived, but still can't understand the meaning of "change in area" mentioned in my photo attached.
Is there any change in area when the conductor is moving across the magnetic field? Does the "change in area" causes any change in magnetic flux, resulting in producing emf( rate of change of magnetic flux) ?

 

[Jy158654]

We say that there are three ways an emf can be induced:
a) changing the magnetic flux density B
b) changing the area A of the circuit
c) changing the angle

Can I know which of the category above is moving the wire/current-carrying conductor across the magnetic field belong to? Why?