A wooden core `( ` figure `)` supports two coils `:` coil `1` with inductance `L_(1)` and shor `-` circuited coil 2 with active resistance `R` and inductance `L_(2)`. The mutual inductance of the coils depends on the distance `x` between them according to the law `L_(12)(x)`. Find the mean `(` averaged over time `)` value of the interaction force between the coils when coil 1 carries an alternating current `I_(1)=I_(0)cos omegat`

Explanation

We consider the force `vec(F_(12))` that a circuit 1 exerts on another closed circuuit 2 `:-`
`vec(F_(12))=ointI_(tau)d vec(l_(2)) xx vec(B_(12))`
Here `vec(B_(12))=` magnetic field at the site of the current element `dvec(l_(2))` due to the current `I_(1)` flowing in 1 .
`=(mu_(0))/( 4 pi ) int(I_(1)dvec(l_(1))xxvec(r_(12)))/(r_(12)^(3))`
where `vec(r_(12))=vec(r_(2))-vec(r_(1))=` vector, current element `dvec(l_(1))` to the current element `dvec(l_(2))`
Now
`vec(F_(12))=(mu_(0))/( 4pi)int intI_(1)I_(2)(dvec(l_(2))xx(dvec(l_(1))xxvec(r_(12))))/(r_(12)^(3))=(mu_(0))/(4pi) int intI_(1)I_(2)(dvec(l_(2))(dvec(l_(2)).vec(r_(12)))-(dvec(l_(1)).dvec(l_(2)))vec(r_(12)))/(r_(12)^(3))`
In the first term, we carry out the integration over `dvec(l_tau)` first. Then
`int int (dvec(l_(1))(dvec(l_(2)).vec(r_(12))))/(r_(12)^(3))=int d vec(l_(2))oint(dvec(l_(2)).vec(r_(12)))/(r_(12)^(3))=-intd vec(l_(1))oint d vec(l_(2)). grad_(2)(1)/( r_(12))=0`
because `oint dl_(2). grad_(2)(1)/(r_(12))=intd vec(S_(2)) curl (grad(1)/(r_(12)))=0`
Thus `F_(12)=-(mu_(0))/(4pi) int int I_(1)I_(2)dvec(l_(1)).dvec(l_(2))(vec(r_(12)))/( r_(12)^(3))`
The integral involved will depend on the vector `vec(a)` that defines the separation of the `(` suitably chosen `)` centre of the coils . Let `C_(1)` and `C_(2)` be the centres of the two coil suitably defined.
Write
`vec(r_(12))=vec(r_(2))-vec(r_(1))=vec(rho_(2))-vec(rho_(1))+vec(a)`
where `vec(rho_(1))(vec(p_(2)))` is the distance of `dvec(l_(1))(dvec(l_(2)))` from `C_(1)(C_(2))` and `vec(a)` stands for the vector `C_(1)vec(C_(2))`.
Then `(vec(r_(12)))/(r_(12)^(3))=-vec(grad_(vec(a)))(1)/(r_(12))`
and `vec(F_(12))=vec("grad")_(a)[I_(1)I_(2)(mu_(0))/( 4pi)int int(dvec(l_(1)).dvec(l_(2)))/(r_(12))]`
The bracket defines the mutual inductance `L_(12)`. Thus noting the definition of `x`
`lt F_(x) gt =(deltaL_(12))/( deltax) lt I_(1)I_(2) gt`
where `lt gt` denotes time average. Now
`I_(1)=I_(2)cos omegat=` Real part of `I_(0)e^(iomegat)` ltbr. The current in coil 2 satisfies `RI_(2)+L_(2)(dI_(2))/(dt)=-L_(12)(dI_(1))/(dt)`
or ` I_(2)=(-iomegaL_(12))/( R+iomegaL_(2))I_(0)e^(iomegat )` ( in the complex case)
taking the real part
`I_(2)=-(omegaL_(12)I_(0))/(R^(2)+omega^(2)L_(2)^(2))( omegaL_(2)cosomegat-R sin omegat)=- (omegaL_(12))/(sqrt(R^(2)+omega^(2)L_(2)^(2)))I_(0)cos ( cosomegat+varphi)`
Where `tan varphi=(R)/( omegaL_(2))`. Taking time average, we get
`lt F_(x) gt =(deltaL_(12))/(delta x)I_(0)(omegaL_(12)I_(0))/(sqrt(R^(2)+omega^(2)L_(2)^(2))).(1)/(2) cos varphi=(omega^(2)L_(2)L_(12)I_(0)^(2))/( 2(R^(2)+omega^(2)L_(2)^(2)))(deltaL_(12))/( deltax)`
The repulsive nature of the force is also consistent with Lenz's law, assuming, of comse, that `L_(12)` decreases with `x`.