emf problem

[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}

[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.

[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.

[Andrew Mason]

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}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

[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

[Andrew Mason]

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}Just a little confused.

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.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

[robert25pl]

Now I understood very well. Thank you very much.