Physics
Doing a study on how mass affects the terminal velocity of a magnet falling through a copper pipe (Magnetic Induction). Need some help!
Step-by-step physics solution: Doing a study on how mass affects the terminal velocity of a magnet falling through a copper pipe (Magnetic Induction). Need
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1. What the question is asking (in plain language)
A magnet is dropped through a copper pipe.
Because the magnet’s field cuts the conducting pipe, eddy currents are induced.
These currents create a magnetic drag that eventually balances the weight of the magnet, so the magnet falls at a terminal velocity (v_{t}).
The student has written the balance as
[ mg = k\,v_{t}, ]
and wants to understand why the constant (k) contains the fourth power of the pipe radius (or diameter).
The goal is to derive the expression for the drag force, show explicitly where the (R^{4}) (or (D^{4})) term comes from, and explain why the experiment should indeed be very sensitive to the pipe size.
2. Step‑by‑step derivation
Below we follow the treatment that appears in the Donoso paper (Eq. 10) but we keep every algebraic step explicit.
2.1. Geometry and notation
| Symbol | Meaning |
|---|---|
| (R) | inner radius of the copper tube (m) |
| (b) | thickness of the tube wall (so outer radius is (R+b)) |
| (\sigma) | electrical conductivity of copper (S · m(^{-1})) |
| (\mu_{0}) | permeability of free space (4π × 10⁻⁷ H · m⁻¹) |
| (a) | radius of the cylindrical magnet (assumed magnetized uniformly along its axis) |
| (L) | length of the magnet (m) |
| (M) | magnetic moment of the magnet (A·m²) – for a uniformly magnetized cylinder (M = \pi a^{2} L\,M_{s}) |
| (v) | instantaneous velocity of the magnet (positive downwards) |
| (g) | gravitational acceleration (9.81 m · s⁻²) |
| (m) | total mass of the falling object (magnet + non‑conductive load) |
The magnet’s axis stays coaxial with the pipe, and the pipe is long enough that end effects can be ignored while the magnet is inside.
2.2. Magnetic field of the moving magnet
For a long cylindrical magnet whose magnetisation is along the axis, the axial component of the magnetic field at a radial distance (r) from the axis (outside the magnet) is well approximated by the field of a magnetic dipole when (r\gg a). Inside the pipe, however, the dominant field component that threads the conducting wall is the axial flux through a circular loop of radius (r).
The flux through a loop of radius (r) at a distance (z) from the centre of the magnet is
[ \Phi(r,z) = \mu_{0} M \frac{z}{2\pi\left(r^{2}+z^{2}\right)^{3/2}} . ]
(Exact expressions exist, but the dipole form already captures the needed (r^{-3}) dependence.)
When the magnet moves with speed (v) the flux changes in time:
[ \frac{d\Phi}{dt}= -v\,\frac{\partial\Phi}{\partial z}. ]
2.3. Induced emf and eddy‑current loop
Consider a thin cylindrical shell of the pipe at radius (r) (with (R\le r\le R+b)) and thickness (\mathrm{d}r).
Because the wall is thin compared with (R) we treat the current as flowing in a circular loop of circumference (2\pi r).
The emf around that loop is
[ \mathcal{E}(r) = -\frac{d\Phi}{dt}= v\,\frac{\partial\Phi}{\partial z}. ]
The resistance of the thin annular strip of width (\mathrm{d}r) and length (2\pi r) is
[ \mathrm{d}R = \frac{1}{\sigma}\,\frac{\mathrm{d}\ell}{A} = \frac{1}{\sigma}\,\frac{2\pi r}{b\,\mathrm{d}r}, ]
where (A = b\,\mathrm{d}r) is the cross‑sectional area of the strip (wall thickness (b) times width (\mathrm{d}r)).
Thus the induced current in the strip is
[ \mathrm{d}I = \frac{\mathcal{E}}{\mathrm{d}R} = \sigma\,\frac{b\,\mathrm{d}r}{2\pi r}\,\mathcal{E}(r). ]
2.4. Lorentz (drag) force on the strip
The current loop experiences a magnetic force density (\mathbf{J}\times\mathbf{B}). Integrating around the loop gives a net axial force (opposite to the motion):
[ \mathrm{d}F = I\, (2\pi r)\, B_{r}(r,z) , ]
where (B_{r}) is the radial component of the magnet’s field at the wall. For a dipole, (B_{r}\sim \dfrac{\mu_{0} M r}{4\pi (r^{2}+z^{2})^{5/2}}).
Putting the expressions together and integrating over the whole wall thickness,
[ \mathrm{d}F = \sigma b \,\frac{\mathcal{E}(r)}{r}\, B_{r}(r,z) \,\mathrm{d}r . ]
Now substitute the explicit forms of (\mathcal{E}) and (B_{r}) (both proportional to (M) and to powers of (r) and (z)). After a bit of algebra the integrand behaves as
[ \mathrm{d}F \propto \sigma b \, v \, M^{2}\, \frac{r^{3}}{(r^{2}+z^{2})^{4}} \,\mathrm{d}r . ]
2.5. Integrate over the pipe wall and over the magnet length
Because the pipe wall is thin we can replace (r) by the mean radius (R) (the error is of order (b/R)). The integral over (r) then simply yields a factor (b) and we obtain
[ F_{\text{drag}} = C\,\sigma\,\frac{M^{2}}{R^{4}}\,v , ]
with
[ C = \frac{\mu_{0}^{2}}{8\pi}\int_{-\infty}^{\infty}\frac{z^{2}}{(R^{2}+z^{2})^{4}}\,\mathrm{d}z = \frac{\mu_{0}^{2}}{8\pi}\,\frac{\pi}{12R^{3}} = \frac{\mu_{0}^{2}}{96R^{3}} . ]
Collecting constants we finally write the linear drag law
[ \boxed{F_{\text{drag}} = k\,v}, \qquad k = \frac{\mu_{0}^{2}\,\sigma\,M^{2}}{96\,R^{4}} . ]
All the approximations (thin wall, dipole field, long pipe) are the same ones used by Donoso; the factor (96) may be written differently depending on how the magnetic moment (M) is expressed. In the paper the authors replace (M) by the product of the magnet’s cross‑sectional area (\pi a^{2}) and its remanent flux density (B_{r}), which yields the same (R^{-4}) dependence.
2.6. Terminal velocity
At terminal velocity the gravitational force equals the magnetic drag:
[ mg = k\,v_{t} \quad\Longrightarrow\quad \boxed{v_{t}= \frac{m g}{k} = \frac{96\, m g\, R^{4}}{\mu_{0}^{2}\,\sigma\,M^{2}} } . ]
Because the drag coefficient (k) contains (R^{4}) (or, equivalently, (D^{4})), a modest change in pipe radius produces a large change in terminal speed.
3. Final answer
The constant that multiplies the velocity in the force balance is
[ \boxed{ k = \dfrac{\mu_{0}^{2}\,\sigma\,M^{2}}{96\,R^{4}} } ]
so that
[ \boxed{ v_{t}= \dfrac{96\, m g\, R^{4}}{\mu_{0}^{2}\,\sigma\,M^{2}} } . ]
The fourth‑power dependence on the pipe radius arises from:
- the induced emf being proportional to the rate of change of magnetic flux ((\propto v/R^{3}) for a dipole field);
- the resistance of the wall scaling as (1/(bR));
- the Lorentz force on the eddy‑current loop bringing in another factor of (1/R).
Multiplying these three contributions gives the overall (R^{-4}) (or (D^{-4})) factor.
Consequently, a pipe whose radius is larger by only a few millimetres can change the drag coefficient – and hence the terminal velocity – by roughly an order of magnitude, exactly as observed in the experiment.
4. Common Mistakes
| Mistake | Why it’s wrong | How to avoid it |
|---|---|---|
| Treating the pipe as a solid cylinder and using its cross‑sectional area for resistance. | The current flows only in the thin wall; the resistance is inversely proportional to the wall thickness, not to the whole area. | Use the wall thickness (b) and the circumference (2\pi R) to compute the resistance of a thin annular strip. |
| Using the magnetic field of a long solenoid instead of the dipole field for a short magnet. | The field decays as (1/r^{3}); using a uniform field underestimates the radial dependence and removes the (R^{-4}) factor. | Approximate the magnet as a dipole (or use the exact analytical expression) when the pipe radius is several times larger than the magnet radius. |
| Neglecting the factor of (v) in the emf and writing the drag force as proportional to (v^{2}). | Faraday’s law gives (\mathcal{E}\propto v); the induced current (and thus the Lorentz force) is linear in (v) for the low‑speed regime where the magnetic diffusion time is short. | Keep the first‑order term in (v); the quadratic term only appears at very high speeds where skin‑effect becomes important. |
| Assuming the magnet’s magnetic moment (M) does not appear in (k). | (M) (or equivalently the product (B_{r} a^{2})) enters squared because the induced emf and the Lorentz force each contain one factor of the field. | Write (M = \pi a^{2}L\,M_{s}) (or (M = B_{r}\,a^{2}L/\mu_{0})) and keep it in the final expression. |
| Forgetting the thin‑wall approximation when the pipe wall is comparable to the radius. | The (R^{-4}) law strictly holds when (b\ll R); a thick wall reduces the effective resistance and changes the exponent slightly. | Verify (b/R < 0.1); otherwise keep the full integral over the wall thickness instead of pulling out a single (R). |
By paying attention to these points the derived expression will match both the literature (Donoso’s Eq. 10) and experimental data, and the strong sensitivity to pipe diameter will be fully understood.
Original question: Doing a study on how mass affects the terminal velocity of a magnet falling through a copper pipe (Magnetic Induction). Need some help! on Physics Stack Exchange, licensed CC BY-SA.