Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 314 / 4 / b

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- Solution b

Solution

The magnetic energy is

E_{B} = \int_{V} \frac{B ^{2}}{2 μ _{0}} d V .

Using the ideal-MHD induction equation and integrating the identity

B \cdot \nabla \times (u \times B)

by parts gives

\frac{d E _{B}}{d t} = \frac{1}{μ _{0}} \int_{S} [(u \times B) \times B] \cdot d S - \int_{V} u \cdot (j \times B) d V .

The volume term vanishes for a force-free magnetic field, leaving

\frac{d E _{B}}{d t} = \frac{1}{μ _{0}} \int_{S} [(u \times B) \times B] \cdot d S .

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