From the Lorentz force law, the Lorentz force on a particle with charge $q$ is given by
$$\vec F = q\left(\vec E + \vec v \times \vec B \right) $$
Clearly, the magnetic force cannot do work on the particle since the magnetic force is always perpendicular to the velocity of the particle.
Now, consider the case that $\nabla \cdot \vec E = 0$ and that the magnetic field is changing with time
$$\vec B = \left( B_0 + \beta t\right) \hat z $$
It follows that
$$\vec E = -\frac{s\beta}{2}\hat \phi $$
For a charged particle constrained to move along a circular path in the $xy$ plane and centered on the $z$ axis, work will be done by this electric field. If the radius of the circular path is $R$, the magnitude of the work done by the electric field on the particle during one revolution is
$$|W_E| = q\pi R \beta $$
The above is simply to show that, through the (non-conservative) induced electric field, energy can be transferred from the magnetic field to a charged particle even though the magnetic force has done no work on the particle.
So, it should not be difficult to see that, in the case of a changing electric current, energy can flow to and from the associated magnetic field via the induced electric field.