Past exam of the advanced fluid dynamics course of the University of Oxford / 2025 / 1 / b

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- Solution b

Solution

Let

E_{B} = \int_{V} B^{2} / (2 μ_{0}) d V

be the total magnetic energy. Multiply the curl form of the resistive induction equation by

B / μ_{0}

and integrate over the fixed volume. The identity

\int_{V} B \cdot \nabla \times C d V = \int_{V} C \cdot \nabla \times B d V + \int_{\partial V} n \cdot (C \times B) d S

follows from the divergence theorem. The stated zero boundary fields remove its surface term. Applying it to the motional and resistive contributions yields the magnetic energy balance in resistive magnetohydrodynamics:

\dot{E}_{B} = \int_{V} (u \times B) \cdot J d V - \frac{η}{μ _{0}^{2}} \int_{V} ∣\nabla \times B ∣^{2} d V = - \int_{V} u \cdot (J \times B) d V - η \int_{V} J^{2} d V .

The first term is the negative of the mechanical power delivered to the fluid by the Lorentz force density. It transfers energy between fluid motion and the magnetic field, with either sign. The second is the nonpositive magnetic contribution to Joule heating. A nonzero positive electrical resistivity therefore dissipates magnetic energy at rate

η \int_{V} J^{2} d V

; it does not force the total magnetic energy to decrease when fluid motion supplies a larger positive power.

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