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Curl of Fields. = S = y. z. Ampere’s Law: integrating the magnetic Field H around a path enclosing the conductor gives the total current in the conductor:- H • dL = I. I. Magnetic Field. H. Curl of Fields. Magnetic Field. H• dL = I x. H• dL I x. =. s. s.
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Curl of Fields = S = y. z Ampere’s Law: integrating the magnetic Field H around a path enclosing the conductor gives the total current in the conductor:- H• dL = I I Magnetic Field H
Curl of Fields Magnetic Field H• dL = Ix H• dL Ix = s s H• dL = curlxH = Jx Lim s 0 s
Curl of Fields Magnetic Field In general where current is flowing in all directions i.e. x,y,z the complete expression for curl H:- curl H = x∂Hz- ∂Hy]+y∂Hx- ∂Hz]+z∂Hy- ∂Hx] ^ ^ ^ ∂y ∂z ∂x ∂x ∂y ∂z ^ ^ ^ curl H = xJx + yJy + zJz = J curl H = x H
Curl of Fields Magnetic Field • Curl H gives the current density Jat a point. • Curl H has a value wherever current is present • DivD gives the charge density at a point. • Div D has a value wherever charge is present. • Curl H is conveniently expressed in vector notation as the cross product of the operator del () and H, that is x H
Curl of Fields Magnetic Field = (x ^ ^ ^ ∂ ∂ ∂ + z y ) + ∂z ∂x ∂y ^ ^ ^ H = xHx + yHy + zHz+ X H = ?
Faraday’s Law Electric Field The Integral of Electric Field E, around a loop of incremental area s enclosing a time changing magnetic flux density B equals the voltage V induced in the loop. V = E•dL = - ∂B•ds = -∂Bs x E =curl E = lim E•dL = - ∂B ∂t ∂t s ∂t s0
curl of Water Flux of water z ^ ^ V = xK = x vx Curl of v has two terms; i.e. ∂vx/ ∂z and ∂vx / ∂y b y x Since vx is constant, hence both terms are zero
curl of Water Flux of water ^ curl vx=xvx = z (K/b cos(y/b) +K/b -K/b b y vx= Ksin(y/b) x ∂vx/∂y = K/b cos(y/b)
Capacitance Flat Plates C = Q/ V Q = sA = DA = EA C =DA/Ed =EA/Ed C = r 0 A/d
Problem: 1 A capacitor has two dielectric media sandwiched as one capacitor, each has same thickness of d= 1cm. Plate area A = 100 cm2 . Neglect the field outside the capacitor. 1=2; 2=3 Find E1, E2, V1, V2, D1, D2 and C