# Emf problem

1. Apr 10, 2005

### robert25pl

A magnetic field is given in the xz-plane by B=Bo/x j Wb/M^2. Consider a rigid square loop situated in the xz-plane with its vertices at (Xo,Zo), (Xo,Zo+b),(Xo+a,Zo+b) and (Xo+a,Zo). If the loop is moving in that plane with the velocity $$V = V_{o}\vec{i}$$ m/s what is the induced emf using Faraday's law

Can someone check my magnetic flux set up

$$\psi=\int_{s}B\cdot\,ds=\int_{x=0}^{x+Vot} \int_{z=0}^{b}\frac{Bo}{x}\vec{j}\cdot\, dx\,dz\vec{j}$$

2. Apr 10, 2005

### StatusX

First find the position of the corners of the loop for any time. Remember, both sides of the loop are moving in the x direction (right now you only have one moving, which means your loop is expanding). Don't forget the corners of the loop are at x0 and z0, not 0 and 0.

3. Apr 10, 2005

### robert25pl

$$\psi=\int_{s}B\cdot\,ds=\int_{xo}^{xo+Vot} \int_{zo}^{zo+Vot}\frac{Bo}{x}\vec{j}\cdot\, dx\,dz\vec{j}$$

My problem is that I don't understand well how to get the position of the corners of the loop for any time because there is not even one example with moving loop in the book.

Last edited: Apr 10, 2005
4. Apr 10, 2005

### Andrew Mason

You want to find the time rate of change of flux so:

Emf $$= \frac{d\phi}{dt} =\frac{d}{dt}\int_{A}B\cdot dA$$

To express flux as a function of t:

$$\phi(x) = \int_{x}^{x + a} b \frac{B_0}{x}dx = bB_0\int_{x}^{x + a} \frac{1}{x}dx = bB_0(ln(\frac{x+a}{x}))$$

Since x = vt:

$$\phi(t) = bB_0(ln(\frac{vt+a}{vt}))$$

So the time rate of change of flux is?...

AM

5. Apr 10, 2005

### robert25pl

So my approach is wrong?

$$\psi=\int_{s}B\cdot\,ds=\int_{xo}^{xo+Vot} \int_{zo}^{zo+Vot}\frac{Bo}{x}\vec{j}\cdot\, dx\,dz\vec{j}$$

and then find emf

$$\oint_{c}E\cdot\,dl=-\frac{d}{dt} \int_{s}B\cdot dS$$

I know my set up of position of corners at any time could be wrong, but why didn't you integrate wrt dz. I understand that the magnetic field is independent of z. Thanks

6. Apr 10, 2005

### Andrew Mason

Just a little confused.

The left side is just the electric potential or emf. The right side is the rate of change of flux (I prefer to use A for area to avoid confusion with Ampere's law.)

To work out the flux through the loop, you have to integrate B over the area. But B is a function of x only, so you can avoid integrating over z by simply letting dA = bdx and integrating over x only. Dividing the area into little strips of length b and width dx and integrating B from x = x to x = x+a gives you the
total flux at a given point.

AM

7. Apr 10, 2005

### robert25pl

Now I understood very well. Thank you very much.