I learnt from book that magnetic field does no work because the force is proportional to $\vec{v}\times\vec{B}$ where $\vec{v}$ is the particle velocity. That vector cross product is always at right angles to $\vec{v}$, so that $\vec{F}\cdot\vec{v}=0$, i.e. no work is done on the particle. But then how come does an inductor store energy in the magnetic field?
Electromagnetism – How an Inductor Stores Energy in the Magnetic Field
electromagnetism
Related Solutions
The question is probably simply asking you to write down the Lorentz force law, rather than rearrange it for $\mathbf{E}$ and $\mathbf{B}$ respectively. You could say: the magnetic flux density is the vector field $\mathbf{B}$ such that the force on a current $\mathbf{I}$ due to it is given by $\mathbf{F} = \mathbf{I} \times \mathbf{B}$.
The cross product, as you say, cannot be inverted. To see this, we note that the direction of $\mathbf{F}$ only tells us that $\mathbf{B}$ must lie in the plane perpendicular to $\mathbf{F}$. Then by the formula:
$$ |\mathbf{F}| = |\mathbf{I}||\mathbf{B}| \sin \theta \,, $$
we see that that the magnitude of $\mathbf{F}$ only pins down the value of $|\mathbf{B}|\sin \theta$, which involves two undetermined quantities. Hence we cannot determine $\mathbf{B}$. If we could invert the cross product, we would do something like this: consider that the cross product is linear, that is:
$$\mathbf{I} \times ( \alpha \mathbf{B}_1 + \beta \mathbf{B}_2) = \alpha \mathbf{I} \times \mathbf{B}_1 + \beta \mathbf{I} \times \mathbf{B}_2\,.$$
This means that we can write our cross product as a matrix equation:
$$ \mathbf{F} = \mathbf{I} \times \mathbf{B} \equiv \mathsf{M} \mathbf{B}\,. $$
Now what is the form of this matrix? To work this out, let's use suffix notation:
$$ F_i = \epsilon_{ijk} I_j B_k \,.$$
So we just have that
$$M_{ik} = \epsilon_{ijk} I_j\,. $$
At this point, you can check that the matrix $\mathsf{M}$ is not invertible, and so we cannot invert to give:
$$ \mathbf{B} = \mathsf{M}^{-1} \mathbf{F} \,,$$
as we would like. Consequently there's simply no way of writing $\mathbf{B}$ in terms of $\mathbf{I}$ and $\mathbf{F}$
A straight wire carrying a current does indeed store energy in a magnetic field so it does have an inductance. For example see Derivation of self-inductance of a long wire.
However the inductance of a straight wire is very small. Coiling the wire into a solenoid allows you to create a circuit element with a large inductance for a small size.
The inductance of a straight wire is given by:
$$ L_\text{wire} = \frac{\mu\ell}{8\pi} \tag{1}$$
The inductance of a coiled wire is normally written in the form:
$$ L_\text{coil} = \frac{\mu N^2 A}{d} $$
where $N$ is the number of turns in the coil, $A$ is the area of the coil and $d$ is the length of the coil. To compare this with equation (1) we note that $A=C^2/4\pi$ where $C$ is the circumference, and $NC$ is the total length of the wire. Substituting this into the equation above gives:
$$ L_\text{coil} = \frac{\mu \ell^2}{4\pi d} \tag{2} $$
Or we for comparison with equation (1) we could use $\ell=N2\pi r$ to rewrite this as:
$$ L_\text{coil} = \frac{\mu \ell}{2}\frac{Nr}{d} \tag{3} $$
Best Answer
From a circuit theory perspective, recall that the product of voltage and current is power:
$p(t) = v(t) \cdot i(t)$
Also, for the inductor:
$v_L(t) = L \dfrac{d}{dt}i_L(t)$
So, there is only a voltage across an inductor when the inductor current is changing with time.
It follows that power (time rate of change of work) is supplied to or delivered from the inductor when the inductor current is changing with time.
But, the magnetic field threading the inductor must be changing with time if the inductor current is changing with time.
Finally, recall that a changing magnetic field induces a non-conservative electric field which can do work.
Remember, for a constant current through an (ideal) inductor, there is no associated power as there is only a steady magnetic field and thus no induced electric field.