Subfield: Electromagnetism

## T1: Solenoid and loop

A closed circular loop of radius $r$ consists of an ideal battery of electromotive force $\mathcal{E}$ and a wire of resistance $R$. A long thin air-core solenoid is aligned with the axis of the loop ( $z$-axis). Its length is $\ell \gg r$, cross-sectional area is $A(\sqrt{A} \ll r)$, and the number of turns is $N$. The solenoid is powered by a constant current $I$ provided by an ideal current source. The directions of the currents in the solenoid and in the loop are the same (clockwise in the figure).

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Question: a. Find the force $F_{1}$ acting on the solenoid when its head $O_{1}$ is positioned in the loop centre $O$. What is the force $F_{2}$ acting on the solenoid when its tail $O_{2}$ is located in the centre of the loop?

Solution:
According to Newton's third law, the force acting on the solenoid is equal in magnitude, but opposite in direction to the force acting on the loop. The latter can be obtained from Lorentz's law by summation of infinitesimal forces $\vec{F}=J \Delta \vec{l} \times \vec{B}$ acting on the individual loop elements $\Delta \vec{l}$, where $J=\mathcal{E} / R$ is the current in the loop.

Since the solenoid is long and thin, the magnetic field lines inside it are directed in $+z$ direction, and escape only from the immediate vicinity of its ends. The magnetic field outside the solenoid is vortex-free and sourcefree. The same requirements are satisfied by the electric field in empty space. Hence, the magnetic field outside the solenoid can be well approximated by a field created by two magnetic poles: a North pole residing close to point $O_{1}$ and a South pole located near to $O_{2}$. The flux $\Phi$ emerging from the North pole (and the flux entering the South pole) is the same as the flux passing through the solenoid's cross-section:

$$
\Phi=B_{\text {in }} A=\mu_{0} \frac{N I}{\ell} A
$$

When the endpoint $O_{1}$ of the solenoid is placed in the loop centre $O$, the magnetic field of the other end $\left(O_{2}\right)$ near the loop is negligible. The field created by the North pole located at $O_{1}$ is pointing radially outwards, and its magnitude at the loop circumference is (from spherical symmetry):

$$
B(r)=\frac{\Phi}{4 \pi r^{2}}=\frac{\mu_{0}}{4 \pi} \frac{N I A}{\ell r^{2}}
$$

Note. it is also correct to assume that the flux from the $B$-field (which carries a factor of $1 / 2$ compared to the field above at the end of the solenoid is uniformly distributed over hemisphere.

The forces acting on all elements of the loop (and also the net force) point in the $-z$ direction, which can be expected also from the same current directions (i.e. the loop and the solenoid attract each other). Hence, the reaction force acting on the solenoid points in the $+z$ direction, and its magnitude is given by:

$$
F_{1}=\frac{\mathcal{E}}{R} \cdot 2 \pi r \cdot \frac{\mu_{0}}{4 \pi} \frac{N I A}{\ell r^{2}}=\frac{\mu_{0} N I A \mathcal{E}}{2 \ell R r} .
$$

When the endpoint $\mathrm{O}_{2}$ is located at the centre of the loop, the magnetic field produced by the South pole exerts a force on the loop. Since this field is directed radially inward, the force acting on the solenoid is the same in magnitude, but opposite in direction $(-z)$ as the force calculated above:

$$
F_2 = -F_1
$$
In this solution the force acting on the solenoid is calculated, as the force acting on the magnetic pole placed at the centre of the current-carrying loop. For this we need to find an expression for the magnetic pole strength (or magnetic charge) $Q_{\mathrm{m}}$, which is defined as the ratio of the force and the magnetic field.

The total dipole moment $m$ of the solenoid is the product of the number of turns and the dipole moment $I A$ of each turn:

$$
m=N I A .
$$

This can be also expressed with the magnetic pole strength and the distance of the poles: $m=Q_{\mathrm{m}} \ell$. From this we arrive to the expression

$$
Q_{\mathrm{m}}=\frac{N I A}{\ell}=\frac{\Phi}{\mu_{0}}
$$

where $\Phi$ is the total flux emerging from the pole (see solution I).

Note. The same result can be obtained from the analogy between electrostatic and magnetostatic fields. The Coulomb force between two point charges $\pm Q$ can be derived from the principle of virtual work. The force is the derivative of the interaction part of the field energy with respect to the distance between the charges. The force between two magnetic charges $\pm Q_{\mathrm{m}}$ can be also calculated this way. From the expressions of electric and magnetic energy densities we can conclude the formula of the magnetic interaction force:

$$
\begin{gathered}
w_{E}=\frac{1}{2} \varepsilon_{0} E^{2} \quad \longleftrightarrow w_{B}=\frac{1}{2 \mu_{0}} B^{2}, \\
F_{E}=\frac{1}{4 \pi \varepsilon_{0}} \frac{Q^{2}}{r^{2}} \quad \longleftrightarrow \quad F_{B}=\frac{\mu_{0}}{4 \pi} \frac{Q_{\mathrm{m}}^{2}}{r^{2}} .
\end{gathered}
$$

As it can be seen, the well-known formulae known in electrostatics can be also used in magnetostatics with the substitutions $\varepsilon_{0}^{-1} \longleftrightarrow \mu_{0}, E \longleftrightarrow B, Q \longleftrightarrow Q_{\mathrm{m}}$. Carrying on this analogy the magnetic pole strength can be figured out:

$$
Q=\varepsilon_{0} \Psi \quad \longleftrightarrow \quad Q_{\mathrm{m}}=\frac{\Phi}{\mu_{0}}=\frac{N}{\ell} I A
$$

where $\Psi$ and $\Phi$ are the electric and magnetic flux for a closed surface containing the electric and magnetic charge, respectively.

When endpoint $O_{1}$ of the solenoid is located at point $O$, a North pole resides at the centre of the loop. Here the magnetic field created by the loop can be expressed from Biot-Savart-law:

$$
B_{\text {loop }}^{(\text {at center })}=\frac{\mu_{0} J}{2 r}=\frac{\mu_{0} \mathcal{E}}{2 R r},
$$



pointing in the $+z$ direction. So the magnitude of the force acting on this end of the solenoid is:

$$
F_{1}=Q_{\mathrm{m}} B_{\text {loop }}^{(\text {at center })}=\frac{\mu_{0} \mathcal{E} N I A}{2 \ell R r}
$$

and it is directed to $+z$. When the endpoint $\mathrm{O}_{2}$ is located at the center of the loop, the force acting on the South pole should be calculated, resulting a force of same magnitude, but opposite direction.
Some of the magnetic field lines created by the loop enter into the near end $O_{1}$ of the solenoid; this entering flux is given by

$$
\Phi_{\text {in }}=B_{\text {loop }}^{\text {(at center) }} A=\frac{\mu_{0} J}{2 r} A=\frac{\mu_{0} \mathcal{E} A}{2 R r} .
$$

Since the end $O_{2}$ is far from the loop, the flux created by the loop escaping there is negligibly small. This means that almost all the flux $\Phi_{\text {in }}$ escapes from the solenoid through its side.

Denote the radial component of the magnetic field vector produced by the current-carrying loop at the perimeter of the $i$ th turn of the solenoid by $B_{i}$. Only this component contributes to the net force acting on the solenoid, as the axial component produces a radial force which is cancelled due to rotational symmetry. The axial force acting on the $i$ th turn of the solenoid is given by

$$
F_{1, i}=2 \sqrt{A \pi} I B_{i}
$$

where $2 \sqrt{A \pi}$ is the circumference of one turn, and the force points in the $+z$ direction. Summing up both sides gives the net force:

$$
F_{1}=\sum_{i} F_{i}=\sum_{i} 2 \sqrt{A \pi} I B_{i}
$$

Take out the factor $I$ from the summation and insert 1 written in the unusual way $(\ell / N) \cdot(N / \ell)$ :

$$
F_{1}=I \frac{N}{\ell} \sum_{i} 2 \sqrt{A \pi} \frac{\ell}{N} B_{i}
$$

The sum on the right hand side is the flux escaping through the side of the solenoid, which equals $\Phi_{i n}$, so the force:

$$
F_{1}=I \frac{N}{\ell} \Phi_{\text {in }}=\frac{\mu_{0} \mathcal{E} A N I}{2 R r \ell}
$$
The force acting on a current loop of magnetic moment $\vec{m}$ placed in magnetic field $\vec{B}$ is given by $\vec{\nabla}(\vec{m} \cdot \vec{B})$. Divide the solenoid into short circular coils of equal length $\Delta \ell$, then the magnetic moment of each short coil is

$$
\Delta \vec{m}=I A \frac{N \Delta \ell}{\ell} \vec{e}_{z}
$$

where $\vec{e}_{z}$ denotes the unit vector in $z$-direction. This magnetic moment is parallel to the field $\vec{B}_{\text {loop }}$ created by the large current-carrying loop, so the force acting on each short segment of the solenoid in $z$-direction can be written as:

$$
\Delta F_{1}=\Delta m \frac{\mathrm{d} B_{\text {loop }}}{\mathrm{d} z}=I A \frac{N \Delta \ell}{\ell} \frac{\mathrm{d} B_{\text {loop }}}{\mathrm{d} z}
$$

The total force on the solenoid can be determined from integration of the force contributions along the solenoid:

$$
F_{1}=\int_{-\ell}^{0} \mathrm{~d} \ell \frac{\Delta F_{1}}{\Delta \ell}=I A \frac{N}{\ell}\left(B_{\text {loop }}(0)-B_{\text {loop }}(-\ell)\right) .
$$

Using the Biot-Savart-law we can compute the magnetic field $B_{\text {loop }}(z)$ of the current-carrying loop along the $z$-axis to

$$
B_{\text {loop }}(z)=\frac{\mu_{0} \mathcal{E}}{2 R} \frac{r^{2}}{\left(z^{2}+r^{2}\right)^{3 / 2}}
$$

This expression for $B_{\text {loop }}$ yields

$$
F_{1}=\frac{\mu_{0} \mathcal{E} A N I r^{2}}{2 R \ell}\left(\frac{1}{r^{3}}-\frac{1}{\left(\ell^{2}+r^{2}\right)^{3 / 2}}\right) \stackrel{\ell}{\approx} r \frac{\mu_{0} \mathcal{E} A N I}{2 \ell R r},
$$

and $F_{1}$ is directed to $+z$. From a similar calculation we get $\vec{F}_{2}=-\vec{F}_{1}$.



Note: Using the idea of dividing the solenoid into small segments other solutions are possible as well (e.g. considering small dipole contributions). One recurring solution of this type is to divide the solenoid into small segments of length $\mathrm{d} z$, compute the $B$-field in $z$-direction effected by current-loop from Biot-Savart, use Gauss law to relate the radial $B$ flux through the segment to $\mathrm{d} B / \mathrm{d} z$. compute the force on the segment using Lorentz forces and integrate the force over the length of the solenoid. In this case the marking scheme is adapted in the second aspect, awarding $0.5 \mathrm{p}$ for the idea of using Gauss law to find radial $B$-field
In this solution we relate the force acting on the solenoid to the change in energy of the system. Investigate the case when point $O_{1}$ is located at $O$ first. Due to the same current directions, the magnetic force $\vec{F}_{1}$ acting on the solenoid points in direction $+z$. While keeping the solenoid in equilibrium with external force $-\vec{F}_{1}$, let it move by a small displacement $\delta z$ in the positive $z$ direction. The work done by the external force is equal to the change in energy of the system:

$$
-\vec{F}_{1} \cdot \delta \vec{z}=-F_{1} \delta z=\delta E_{\text {total }}
$$

We should be aware of the fact that the system is not closed: there is also a battery and a current source included in the circuits. Hence, $\delta E_{\text {total }}$ contains the change in field energy and the change of energy of the power sources:

$$
\delta E_{\text {total }}=\delta E_{\text {field }}+\delta E_{\text {sources }}
$$

Since the force does not depend on what kind of power supplies we have, let us replace the battery with a current source providing constant current $J=\mathcal{E} / R$.

Now we find a relation between $\delta E_{\text {field }}$ and $\delta E_{\text {sources }}$ The energy stored in the field can be expressed as

$$
E_{\text {field }}=\frac{1}{2} L_{1} I^{2}+\frac{1}{2} L_{2} J^{2}+L_{12} I J
$$

where $L_{1}$ is the inductance of the solenoid, $L_{2}$ is that of the loop and $L_{12}$ is the mutual inductance of the system. Upon small displacement $\delta z$, only the last term changes, so

$$
\delta E_{\text {field }}=\delta L_{12} \cdot I J
$$

The small displacement results in a change of the flux enclosed by the loop and the solenoid. The flux created by the solenoid on the loop is $L_{12} I$, and the flux created by the loop through the solenoid is $L_{21} J=L_{12} J$ (here we used the symmetry property of mutual inductance). During the short time $\delta t$ of the displacement $\delta z$, the e.m.f. induced in the loop ( $\left.V_{\text {ind }}^{\text {loop }}\right)$ and the solenoid $\left(V_{\text {ind }}^{\text {solenoid }}\right)$ can be expressed with Faraday's law:

$$
V_{\text {ind }}^{\text {loop }}=-\frac{\delta L_{12}}{\delta t} I, \quad V_{\text {ind }}^{\text {solenoid }}=-\frac{\delta L_{12}}{\delta t} J .
$$

In order to keep the current in the circuits constant, the current sources need to provide an additional power, so they give away extra energy (in addition to Joule heat). This energy change of the sources is given by:

$$
\delta E_{\text {sources }}=\left(V_{\text {ind }}^{\text {loop }} J+V_{\text {ind }}^{\text {solenoid }} I\right) \delta t
$$

Using the previous results we finally get:

$$
\delta E_{\text {sources }}=-2 \delta L_{12} I J=-2 \delta E_{\text {field }}
$$

which means $\delta E_{\text {total }}=-\delta E_{\text {field }}$, and hence $F_{1} \delta z=$ $\delta E_{\text {field }}=\delta L_{12} I J$.

Note 1. Naively, one might think that we get the result $\delta E_{\text {total }}=\delta E_{\text {field }}$ if we imagine superconducting wires without power supplies. One can show with detailed calculation that in that case the currents in the loop and the coil change, as the total flux enclosed by a superconducting circuit must remain constant. The correct physical justification of the appearing negative sign is an important part of the solution.

Now we calculate the change in mutual inductance $\delta L_{12}$. The small displacement $\delta z$ can be imagined as we take a short segment from the tail $\mathrm{O}_{2}$ of the coil (consisting of $N \delta z / \ell$ turns) and move it to the head $O_{1}$. As a result, the flux produced by the loop on the solenoid increases by

$$
\delta \Phi_{12}=\delta L_{21} J=\underbrace{\frac{\mu_{0} J}{2 r}}_{\substack{\text { (at center) } \\ \text { loop }}} A \frac{N}{\ell} \delta z
$$

From this we get:

$$
F_{1}=\frac{\delta L_{12} I J}{\delta z}=\frac{\mu_{0} \mathcal{E} N A I}{2 \ell R r}
$$

If the tail $\mathrm{O}_{2}$ of the solenoid is located at point $O$, the coefficient of mutual inductance decreases upon small displacement, which results in $\vec{F}_{2}=-\vec{F}_{1}$.

Note 2. The field energy can be also calculated from the energy density integrated for the whole space. Instead of calculating the total field energy, it is easier to find its change using the same idea presented above, i.e. take a segment of length $\delta z$ from the tail and move it to the head of the solenoid. Assuming $\ell \gg \delta z \gg \sqrt{A}$, the field created by the solenoid inside that segment is $\mu_{0} N I / \ell$ (because the field differs from this only at distance $\sim \sqrt{A}$ from the ends). At the end we get the same result for the change in field energy using the expression:

$$
\delta E_{\text {field }}=\frac{1}{2 \mu_{0}}\left[\left(B_{\text {loop }}^{\text {(at centre) }}+B_{\text {sol }}\right)^{2}-\left(B_{\text {loop }}^{\text {(at centre) })}\right)^{2}-B_{\text {sol }}^{2}\right] A \delta z .
$$

Note 3. A third possibility is to calculate the potential energy change of the displaced few turns of the solenoid. The magnetic moment of a segment of length $\delta z$ is $\vec{m}=\vec{e}_{z} I A N \delta z / \ell$, and its energy in external field is $E_{\mathrm{pot}}=-\vec{m} \vec{B}$. Important to highlight that this potential energy already contains the factor of -1 discussed at the beginning of the solution, so the force acting on the solenoid can be expressed as

$$
F_{1}=-\frac{\delta E_{\mathrm{pot}}}{\delta z}
$$

The external field is the superposition of the field $\vec{B}_{\text {loop }}$ created by the loop and the field $\vec{B}_{\text {sol }}$ created by the coil (note that this latter contains a factor of $1 / 2$ compared to the field in the middle of the solenoid). Since $\vec{B}_{\text {sol }}$ is the same at the two ends $O_{1}$ and $O_{2}$, the energy change is:

$$
\delta E_{\mathrm{pot}}=-\vec{m} \vec{B}_{\text {loop }}^{\text {(at center) }}-\vec{m} \vec{B}_{\text {loop }}(z=\ell) .
$$

The second term can be neglected, and we get

$$
\delta E_{\mathrm{pot}}=-\frac{I A N \delta z}{\ell} \frac{\mu_{0} \mathcal{E}}{2 R r}
$$



which gives the same answer for $F_{1}$ as the other ideas.

Final answer: $F_{1}=\frac{\mu_{0} N I A \mathcal{E}}{2 \ell R r}$, $F_2=-F_1$