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Chapter 1

P1.2 a) e1 = cos φ e1 + sin φ e2, e2 = −sin φ e1 + cos φ e2, e3 = e3

A′1 = A1 cos φ + A2 sin φ, A′2 = − A1 sin φ + A2 cos φ, A′3 = A3

b) B = (μoI/2πr2)(−yex + xey) is a vector equal to (μoIr2)( ez × r′), while B′ = (μoI/2πr2) [(x′ sin 2φ + y′ cos 2φ) e1 + (x′ cos 2φ − y′ sin 2φ) e2] is not a vector.

P1.4 n.E(r) = 4πR2f(R), ∇.E = 2f/r + ∂rf and ∇.E = 4πR2f(R).

P1.5 Let xα = Σβ Rβα xβ. We find ∂α = Σβ ∂′β.(∂xβ/∂xα) = Σβ ∂′β Rαβ = Σβ Rαβ ∂′β.

P1.6 b) If dr is on V = Constant, dV = ΣααV dxα = V.dr = 0, hence ∇V is normal to dr. c) If only x varies, as r2 = x2 + y2 + z2, we get 2r dr = 2x dx, hence ∂αr = xα/r and ∂αf(r) = (df/dr)(∂αr) = (df/dr)(xα/r). If f = K/r, we find ∂αf(r) = (−K/r2)(xα/r) = −Kxα/r3.

P1.8 a) If V = K(p.r)/r3, E = −V = (K/r3)[3(p.r) r/r2p]. If p = pez, V = Kpz/r3 and E = (Kp/r3)[3zr/r2ez]. b) If , we find .Using spherical coordinates, if , we find sin θ eφ/r2 and .

P1.9 As ∇ × E = 0, E is the gradient of ...

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