Electricity and Magnetism · Problem 7.18

ALTERNATING-CURRENT CIRCUITS — Edward Purcell & David Morin

Physics — E&M · magnetic-energy-two-expressions-general

Problem

Two expressions for the energy (general) *** The task of Problem 2.24 was to demonstrate that two different expressions for the electrostatic energy, ∫(ϵ0E2/2) dv\int(\epsilon_0 E^2/2)\,dv and ∫(ρϕ/2) dv\int(\rho\phi/2)\,dv, are equivalent (as they must be, if they are both valid). The latter expression can quickly be converted to Cϕ2/2C\phi^2/2 in the case of oppositely charged conductors in a capacitor (see Exercise 3.65). The task of this problem is to demonstrate the analogous relation for the magnetic energy, that is, to show that if a circuit (of finite extent) with self-inductance LL contains current II, then ∫(B2/2μ0) dv\int(B^2/2\mu_0)\,dv equals LI2/2LI^2/2. This is a bit trickier than the electrostatic case, so here are some hints: (1) a useful vector identity is ∇⋅(A×B)=B⋅(∇×A)−A⋅(∇×B)\nabla \cdot (\mathbf{A} \times \mathbf{B}) = \mathbf{B} \cdot (\nabla \times \mathbf{A}) - \mathbf{A} \cdot (\nabla \times \mathbf{B}), (2) the vector potential and magnetic field satisfy ∇×A=B\nabla \times \mathbf{A} = \mathbf{B}, (3) ∇×B=μ0J\nabla \times \mathbf{B} = \mu_0 \mathbf{J}, (4) Φ=∫A⋅dl\Phi = \int \mathbf{A} \cdot d\mathbf{l} from Eq. (7.52), and (5) LL is defined by Φ=LI\Phi = LI.

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