Classical electrodynamics - Klassische Elektrodynamik by Wegner F.

By Wegner F.

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E. there is no magnetic monopole. With g(r) = xα xβ one obtains d3 r xα jβ (r) + xβ jα (r) = 0. 5) Thus we can rewrite 1 2 d3 r(xα jβ − xβ jα ) + 1 2 d3 r(xα jβ + xβ jα ). 6) Das zweite Integral verschwindet, wie wir gerade The second integral vanishes, as we have seen. The gesehen haben. Das erste a¨ ndert sein Vorzeichen bei first one changes its sign upon exchanging the indices Austausch der Indices α und β. Man f¨uhrt ein α and β. 8) 52 C Magnetostatik C Magnetostatics als das magnetische Dipolmoment.

Die elektrischen Potentiale Φi der Leiter #i seien vorgegeben. Gesucht sind die freibeweglichen Ladungen qi auf den Leitern. Da die MGleichungen linear sind (und wir annehmen, dass lineare Beziehungen D = E bestehen), k¨onnen wir das Potential als Superposition von L¨osungen Ψi schreiben Φ(r) = Φi Ψi (r). 9) i Dabei ist Ψi die L¨osung, die auf dem Leiter #i den Wert 1, auf den anderen den Wert 0 annimmt Ψi (r) = δi, j Ψi is the solution which assumes the value 1 at the conductor #i, and 0 at all others Leiter r∈ j.

0023 and the components of the spin s are ± /2. 927 · 10−20 dyn1/2 cm2 . α Force and Torque on a Dipole in an External Magnetic Field Force Eine a¨ ußere magnetische Induktion Ba u¨ bt auf einen An external magnetic induction Ba exerts on a loop of Ringstrom die L-Kraft a current the L force 1 1 ∂Ba ∂Ba 1 d3 rj(r) × Ba (r) = − Ba (0) × d3 rj(r) − × d3 rxα jβ (r)eβ − ... 19) K= c c c ∂xα ∂xα aus. Wir formen mγ α,β,γ eβ = mγ eγ × eα = m × eα um . We rewrite mγ α,β,γ eβ = mγ eγ × eα = m × eα and find und finden ∂Ba ∂Ba ∂Ba K=− × (m × eα ) = (m · )eα − (eα · )m.

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