Deriving the Force on an Infinitessimal current-loop in a Magnetic Field

In summary, the conversation discusses deriving an equation by evaluating an integral along a square loop. The solution involves using Taylor expansion and evaluating the integral, but there is confusion about a step in the solution. Specifically, the solution manual states that an expression can be approximated, but it is not clear why this approximation is valid.
  • #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?
 
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  • #2
OGrowli said:
[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]

[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]

Does anyone know what is wrong with the LaTex coding here?
 

Related to Deriving the Force on an Infinitessimal current-loop in a Magnetic Field

1. What is a current-loop in a magnetic field?

A current-loop is a closed path or circuit through which electric current can flow. When placed in a magnetic field, the loop experiences a force due to the interaction between the magnetic field and the moving charges within the loop.

2. What is the force on an infinitesimal current-loop in a magnetic field?

The force on an infinitesimal current-loop in a magnetic field is given by the equation F = I∆l x B, where I is the current flowing through the loop, ∆l is the length of the loop, and B is the magnetic field strength. This force is perpendicular to both the current and the magnetic field, and its magnitude is proportional to the current and the magnetic field strength.

3. How is the direction of the force on an infinitesimal current-loop determined?

The direction of the force on an infinitesimal current-loop is determined by the right-hand rule. If you point the fingers of your right hand in the direction of the current and curl them towards the magnetic field, your thumb will point in the direction of the force.

4. What is the significance of using an infinitesimal current-loop in this equation?

An infinitesimal current-loop is used in this equation because it simplifies the calculation of the force. By considering an infinitely small loop, we can apply the equation to any point within the loop and then integrate over the entire loop to find the total force on the loop.

5. How is this equation used in practical applications?

This equation is used in many practical applications, such as designing motors and generators, measuring magnetic field strength, and understanding the behavior of charged particles in a magnetic field. It is also a fundamental concept in the study of electromagnetism and is essential for understanding the relationship between electricity and magnetism.

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