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There is a conducting circular loop with resistance R falling in the magnetic field \vec{B}=B_\rho(\rho,z)\hat\rho+B_z(\rho,z)\hat z and gravitational field \vec g=-g\hat z. How does z and the current in the loop change in time?(assume the loop remains horizontal!)
The flux through the loop is \int_0^a\int_0^{2\pi} B_z \rho d\varphi d\rho, Its time derivative is \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho and so the induced current is I=\frac{1}{R} \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho. Now we can write the z component of the magnetic force as -\frac{2\pi a B_\rho}{R} \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho. So we have:
<br /> <br /> \ddot z=-g-\frac{dz}{dt}\frac{4\pi^2 a B_\rho}{m R} \int_0^a \frac{\partial B_z}{\partial z} \rho d\rho<br /> <br />
Which gives us z as a function of time and then I can be calculated easily.
1- Is everything OK?
2-Any hints or suggestions or further explanations?
3-How does this change if the loop is superconducting?
Thanks
The flux through the loop is \int_0^a\int_0^{2\pi} B_z \rho d\varphi d\rho, Its time derivative is \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho and so the induced current is I=\frac{1}{R} \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho. Now we can write the z component of the magnetic force as -\frac{2\pi a B_\rho}{R} \int_0^a \int_0^{2\pi} \frac{\partial B_z}{\partial z} \frac{dz}{dt} \rho d\varphi d\rho. So we have:
<br /> <br /> \ddot z=-g-\frac{dz}{dt}\frac{4\pi^2 a B_\rho}{m R} \int_0^a \frac{\partial B_z}{\partial z} \rho d\rho<br /> <br />
Which gives us z as a function of time and then I can be calculated easily.
1- Is everything OK?
2-Any hints or suggestions or further explanations?
3-How does this change if the loop is superconducting?
Thanks
