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Curl of Fields

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

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  1. 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

  2. Curl of Fields Magnetic Field  H• dL = Ix  H• dL Ix = s s  H• dL = curlxH = Jx Lim s 0 s

  3. 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

  4. 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

  5. Curl of Fields Magnetic Field  = (x ^ ^ ^ ∂ ∂ ∂ + z y ) + ∂z ∂x ∂y ^ ^ ^ H = xHx + yHy + zHz+  X H = ?

  6. 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 = -∂Bs  x E =curl E = lim  E•dL = - ∂B ∂t ∂t s ∂t s0

  7. 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

  8. 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)

  9. Capacitance Flat Plates C = Q/ V Q = sA = DA = EA C =DA/Ed =EA/Ed C = r 0 A/d

  10. 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

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