Past exam of the physics 2 paper of the University of Oxford / 2013 resit / 9

Let $A_t=\pi d^2/4$ be the tube cross-section. A long uniformly wound solenoid with $n_1$ total turns has field $B_A={\mu }_0{n}_{1}I/s$, approximately uniform inside and negligible outside. Therefore the flux through one cross-section, or one turn of either coil, isThe linked flux of coil A is $n_1\Phi$; that of coil B is $n_2\Phi$. Distinguishing flux from flux linkage prevents an extra factor of the turn count.

For chosen positive terminal currents, define self-inductance by the coil's own flux linkage ${\Lambda }_j=L_j{I}_j$ and mutual inductance by ${\Lambda }_2=M{I}_1$ when only coil A carries current. With matching winding directions and terminal references,The reciprocal flux-linkage relation gives the same $M$ from B to A, and $M^2=L_A{L}_B$ in this ideal fully coupled long-solenoid limit. The induced emf is $-d\Lambda /dt$; passive terminal voltage drops use the opposite sign, $v_A=L_A{\dot{I}}_A+M{\dot{I}}_B$ and similarly for B.

Reversing one winding reverses its signed coupling to the other, while its self-inductance is unchanged. With the same north-to-south current references,Equal currents then produce opposing magnetic fields: the signed axial field is proportional to $(n_1-n_2)I$, rather than $(n_1+n_2)I$. Which direction is positive must be fixed by the right-hand rule and the actual winding orientation.

Let $B_E>0$ be the northward component of the ambient magnetic field. Define $\eta =+1$ if positive north-to-south current in A produces a northward field, and $\eta =-1$ otherwise. With opposite windings and the same signed current in both,A compass reverses only when this northward component becomes negative. For current chosen to make the coil field southward,At equality the ideal axial field is zero and the needle has no preferred axial direction. If $n_1=n_2$, the opposing coil fields cancel and no finite current reverses it in this model. A numerical threshold cannot be given without $B_E$ and the coil parameters.

Use north-minus-south terminal voltages for both oppositely wound coils, with coil A measured at high input impedance so its AC current is zero. Then $v_B=L_B{\dot{I}}_B$ and $v_A=M{\dot{I}}_B$, with $M<0$. HenceThe magnitude is the turns ratio times the drive, and these terminal references give a phase reversal. Reversing one voltmeter lead reverses the displayed sign. A fixed DC bias in A does not change this incremental result, while short-circuiting A would require solving both coupled current equations.