Answer
$-0.00005$ amp/s
Work Step by Step
As we know that $V=IR$
and $\dfrac{dV}{dt}=R\dfrac{dI}{dt}+I\dfrac{d R}{d t}$
Plug the given values respectively.
$(-0.01)=(600)\dfrac{dI}{dt}+(0.04)(0.5)$
$\implies \dfrac{dI}{dt}=\dfrac{(-0.01)-(0.04)(0.5)}{600}$
Hence, $\dfrac{dI}{dt}=-0.00005$ amp/s
