Past exam of the physics B course of the University of Cambridge / 2021 / Paper 1 / Section C / C9

For a steady filamentary electric current $I$ in vacuum, the Biot-Savart law isThe line element points with conventional electric current, and the cross product fixes the magnetic flux density direction by the right-hand rule.

As an application, an infinitely long straight wire has circular symmetry about its axis. At perpendicular distance $s$,Its direction follows the right-hand rule. The following part applies the same law to a circular current loop.

Parameterize the loop by ${\mathbf{r}}^{\prime }=(a\cos \varphi ,a\sin \varphi ,0)$ and choose positive electric current counterclockwise as seen from $+z$. At the observation point $\mathbf{r}=(0,0,z)$,The numerator of the Biot-Savart law isThe distance $\sqrt{a^2+z^2}$ is constant round the loop. The transverse terms integrate to zero, while the axial term integrates to $2\pi a^2$. HenceFor use below, define the undisturbed magnetic field intensity $H_0(z)=B_0(z)/{\mu }_0=a^{2}I/[2(a^2+z^2{)}^{3/2}]$. This distinguishes the applied field from the internal field after adding the magnetic sample.

The printed formula has a physical problem for the specified geometry. Let $\chi =\mu -1$ be the linear magnetic susceptibility, where $\mu$ is the source's relative magnetic permeability. The constitutive equation is $M=\chi H_{\mathrm{in}}$, not in general $\chi H_0/\mu$.

For a small uniformly magnetized ellipsoid with its principal axis along the applied field, include its demagnetizing field through the SI demagnetizing factor $N_d$:Solving givesA slender rod aligned along the field has $N_d\to 0$ as its aspect ratio grows. Its ideal slender-rod limit at fixed $\mu$ is thereforeprovided $N_d|\mu -1|\ll 1$. A finite cylinder generally has nonuniform magnetization, so an ellipsoidal shape factor is an approximation, and its aspect ratio is needed for a more precise correction. Small sample length compared with the field-variation scale makes $H_0$ approximately uniform across it; that condition does not set the demagnetizing factor to one.

The source's displayed $M=(\mu -1)H_0/\mu$ instead follows from $N_d=1$, the thin-plate limit with field normal to the plate. It can also be obtained by imposing $B_{\mathrm{in}}=B_0$, but that identification is not justified for a thin rod parallel to the field. Consequently the requested extra $1/\mu$ cannot honestly be derived from the stated rod geometry. The next part gives both the shape-consistent result and the force conditional on using the printed magnetisation.

In the small-body approximation, write ${\chi }_{\mathrm{eff}}=\chi /(1+N_d\chi )$, with $\chi =\mu -1$. The induced magnetic dipole moment is $\mathbf{m}=VM\hat{\mathbf{z}}=V{\chi }_{\mathrm{eff}}{H}_0\hat{\mathbf{z}}$. Its effective interaction potential energy at fixed source current isThe factor $1/2$ accounts for inducing the moment: differentiating $-m(z)B_0(z)$ as if the moment were permanent would double the force. Equivalently, the instantaneous dipole force is $F_z=md{B}_0/dz$. Thus the force on a small linearly magnetizable body isFrom part (b), $d{H}_0/dz=-3a^{2}Iz/[2(a^2+z^2{)}^{5/2}]$. HenceFor the ideal slender rod, ${\chi }_{\mathrm{eff}}\simeq \mu -1$. If the source's printed magnetisation is imposed instead, its corresponding conditional answer isFor $\mu >1$ and $z>0$, both expressions point toward the loop, where the applied magnetic field is stronger; for $z<0$ the direction reverses toward the loop plane. A material with diamagnetism, ${\chi }_{\mathrm{eff}}<0$, reverses this attraction. The force is in newtons, since a magnetic dipole moment times a field gradient has units $\mathrm{A}{\mathrm{m}}^2\mathrm{T}{\mathrm{m}}^{-1}$. Both formulas assume linear response, negligible saturation and approximately uniform applied field across the sample.

For a linear induced response, reversing $I$ reverses $H_0$ and the sample magnetization, while also reversing the field gradient. Their product in $F_z={\mu }_{0}V{\chi }_{\mathrm{eff}}{H}_0{H}_0^{\prime }$ therefore keeps the same sign and magnitude:This is also explicit in the $I^2$ dependence in part (d), for both the shape-consistent and printed alternatives. The conclusion assumes no independent permanent moment, remanence or hysteresis; those would require a different response model.

Changing shape changes the demagnetizing field, and hence the magnetization induced by the same applied magnetic field intensity. For a small linear sample in the ellipsoid approximation, the force magnitude scales withFor $\mu >1$, a more elongated body aligned with the field has smaller $N_d$ and a stronger response. A sphere has $N_d=1/3$, giving ${\chi }_{\mathrm{eff}}=3(\mu -1)/(\mu +2)$; a very thin plate with field normal to it has $N_d\to 1$, giving $(\mu -1)/\mu$. Thus an axially slender rod generally experiences a stronger attraction than a sphere or a transverse thin plate of the same material and volume, within the linear uniform-applied-field approximation.

The shape response is precisely why the $1/\mu$ factor in part (c) cannot apply universally to every small magnetic body. Orientation also matters through the appropriate principal demagnetizing factor; for general nonellipsoidal bodies the internal field is not uniform and a single scalar correction need not be exact.