# Induced Current, Dipole Falling Under Gravity Towards Circular Loop [closed]

Suppose a magnetic dipole $\mathbf{m} = m \hat{z}$ is falling towards a circular loop of radius $b$ under gravity. Assuming the dipole always stays along the $z$-axis of the loop, determine the following:

In terms of the height $z$, loop resistance $R$, and speed $v$, determine

(a) The EMF induced around the loop.

(b) The magnetic field of the loop at the dipole created by the induced current.

(c) The force on the dipole.

So far, I have found that flux through the loop of radius $b$ is $\phi = \frac{\mu_0 m b^2}{2(b^2 + z^2)^{3/2}}$ from treating the dipole as a current loop with radius $\epsilon$, where $b >> \epsilon$.

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## closed as too localized by John Rennie, twistor59, akhmeteli, Manishearth♦May 5 '13 at 14:03

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Assume that the loop lies on the x-y plane.You can solve this problem either in the general case ($z$ can have any value) or for the case of $z>>b$ where you can assume the field of dipole is uniform on the surface of the loop. We solve for the general case.

Now we find the magnetic field of the dipole on this plane ($z=0$).We use cylindrical coordinates: (I hope I don't make an algebraic mistake!) $$\mathbf{B}=\frac{\mu_0}{4\pi}\frac{3(\mathbf m \cdot \hat{r})\hat{r}-\mathbf m}{r^3}$$

$$\mathbf m=m \hat z$$ $$\hat{r}=\frac{\rho \hat{\rho}-z\hat{z}}{\sqrt {z^2+\rho^2}},r=\sqrt {{z^2+\rho^2}}$$

Now we want to find the flux into the loop.

$$\phi=\int_{loop} \mathbf{B}\cdot d\mathbf{S}$$ $$d\mathbf{S}=2\pi\rho d\rho \hat{z}$$

Substituting this in the relation for $B$ you will have:

$$\phi=\frac{\mu_0 m}{2} \int_0^b \frac{3z}{(z^2+\rho^2)^2}( 1+\frac{z}{\sqrt {z^2+\rho^2}})\rho d\rho$$ This integral can be evaluated as: $$\phi=\frac{-\mu_0m}{2}z (\frac{2z+3\sqrt {z^2+b^2}}{2(z^2+b^2)^\frac{3}{2}}-\frac{5}{2z^2})$$ for the sake of brevity we use $A(z)$ instead of the expression in the parenthesis: $$A(z)=z (\frac{2z+3\sqrt {z^2+b^2}}{2(z^2+b^2)^\frac{3}{2}}-\frac{5}{2z^2})$$ so $$\phi=\frac{-\mu_0m}{2}A(z)$$

now we find time derivative of this: $$\mathcal{E}=-\frac{d\phi}{dt}=-\frac{d\phi}{dz}\frac{dz}{dt}=-\frac{d\phi}{dz}\dot z$$ after finding this derivative, we will find induced $I$: $$I={\mathcal{E} \over R} \to I=\frac{\mu_0 m}{2b} {dA(z) \over dz} \dot z$$ now we find the magnetic field due to this current on the $z$ axis (at the location of the dipole) $$\mathbf{B}=\frac{\mu_0Ib^2}{2(b^2+z^2)^\frac{3}{2}} \hat z=\frac{\mu_0^2bmA(z)}{4(b^2+z^2)^\frac{3}{2}} \dot z \hat z$$ Again we use the following substitution for simplicity: $$G(z)=\frac{A(z)}{(b^2+z^2)^\frac{3}{2}}$$ so $$\mathbf{B}=\frac{\mu_0^2bmG(z)}{4} \dot z \hat z$$

and $$\dot z=v_z$$ The exerted force will be

$$\mathbf{F}=\nabla ({m \hat z \cdot \mathbf B})= \hat z m{d \over dz}B$$ $$\mathbf{F}=\hat z \frac{\mu_0^2bm^2}{4} ({dG \over dz} v_z+ G{dv_z \over dz})$$

now we write ${dv_z \over dz}$ as follows :( assume $a={dv_z \over dt}$ is acceleration) $${dv_z \over dt}=v_z{dv_z \over dz} \to {dv_z \over dz}={1 \over v_z}{dv_z \over dt} \to {dv_z \over dz}={a \over v_z}$$

assume that the mass of the loop is $M$ .Now we write the equation of motion : $$M a \hat z=-Mg \hat z +\hat z \frac{\mu_0^2bm^2}{4} ({dG \over dz} v_z+ G{a \over v_z} )$$

So you will find acceleration as:(we omit $z$ index from $v_z$ ) $$a=\frac{-Mg +\frac{\mu_0^2m^2bv}{4} {dG \over dz}}{M-\frac{\mu_0^2m^2bG}{4v} }$$ and $$\mathbf{F}=Ma\hat z$$

If you solve the problem for the simpler case of $z>>b$ , it is easy to show that you will have the following for the flux:

$$\phi=B(z=0,\rho =0) \pi b^2=\frac{\mu_0mb^2}{2z^3}$$

In this case just substitute the following for $A(z)$ and repeat the steps:

$$A'(z)={-b^2 \over z^3}$$ (as expected, you will arrive at this expression from above $A(z)$ for $b<<z$ if you expand it to second order with respect to $b \over z$ .)

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