- #1
OGrowli
- 14
- 0
We're asked to derive the following equation:
[tex] F=\triangledown(\vec{m}\cdot \vec{B}) [/tex]
by evaluating [tex] F=\int I(d\vec{l}\times \vec{B})[/tex] along a sqaure loop with sides of length ε, parralel to the yz plane. The square's bottom left corner is situated at the origin.
so far I have,
[itex] d\vec{l} \times \vec{B}= -dz \times \hat{z} \times \vec{B}(0,0,z) + dy \hat{y} \times \vec{B}(0,y,0)+dz\hat{z} \times \vec{B}(0,\epsilon ,z)-dy \hat{y} \times \vec{B}(0,y,\epsilon) [/itex]
then using taylor expansion to turn B(0,y,ε) and B(0,ε,z) into expressions of B(0,y,0) and B(0,0,z) respectively and combining like terms I've got:
[itex]
- \hat{y} \times \left \epsilon dy\frac{\partial \vec{B}} {\partial z} \right |_{0,y,0} + \hat {z} \times \left \epsilon dz \frac{\partial \vec{B}} {\partial y} \right |_{0,0,z}
[/itex]
the next step would be to evaluate ∫dF and the solutions manual says that,
[itex]
\left \int dz\frac{\partial\vec{B}}{\partial y} \right |_{0,0,z}\approx \left \epsilon \frac{\partial\vec{B}}{\partial y} \right |_{0,0,0}
[/itex]
from this point the solution comes pretty easily, but I still don't understand why the above is true. Can anyone explain this to me?
[tex] F=\triangledown(\vec{m}\cdot \vec{B}) [/tex]
by evaluating [tex] F=\int I(d\vec{l}\times \vec{B})[/tex] along a sqaure loop with sides of length ε, parralel to the yz plane. The square's bottom left corner is situated at the origin.
so far I have,
[itex] d\vec{l} \times \vec{B}= -dz \times \hat{z} \times \vec{B}(0,0,z) + dy \hat{y} \times \vec{B}(0,y,0)+dz\hat{z} \times \vec{B}(0,\epsilon ,z)-dy \hat{y} \times \vec{B}(0,y,\epsilon) [/itex]
then using taylor expansion to turn B(0,y,ε) and B(0,ε,z) into expressions of B(0,y,0) and B(0,0,z) respectively and combining like terms I've got:
[itex]
- \hat{y} \times \left \epsilon dy\frac{\partial \vec{B}} {\partial z} \right |_{0,y,0} + \hat {z} \times \left \epsilon dz \frac{\partial \vec{B}} {\partial y} \right |_{0,0,z}
[/itex]
the next step would be to evaluate ∫dF and the solutions manual says that,
[itex]
\left \int dz\frac{\partial\vec{B}}{\partial y} \right |_{0,0,z}\approx \left \epsilon \frac{\partial\vec{B}}{\partial y} \right |_{0,0,0}
[/itex]
from this point the solution comes pretty easily, but I still don't understand why the above is true. Can anyone explain this to me?