Answer
$
\Delta U=5.6×10^{-10}eV$
Work Step by Step
(b) The energy required to turn it end-for-end (from $\phi=0^{\circ}$ to $\phi=180^{\circ}$ ) is
$
\Delta U=2 \mu B=2\left\{15 \times 10^{-23} \mathrm{~J} / \mathrm{T} \mid\left[3.0 \times 10^{-6} \mathrm{~T} \mid \mathrm{h}=9.0 \times 10^{-29} \mathrm{~J}=5.6 \times 10^{-10} \mathrm{eV}\right.\right.
$
