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A wooden core (figure ) supports two coils : coil 1 with inductance L_(1) and shor - circuited coil2 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 |
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Answer» Solution :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`. |
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