Q.Consider a wire carrying a steady current, $I$ placed in a uniform magnetic field $\vec{B}$ perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that,

A magnetic force is always $\vec{F}=q(\vec{v}\times\vec{B})$, perpendicular to the charge's own velocity $\vec{v}$, so $dW=\vec{F}\cdot\vec{v}\,dt=0$ - for any charge, however that velocity arose.

1. If the wire itself is set moving by the field, the lattice ions now carry the wire's velocity too - but the magnetic force is still perpendicular to that velocity, so it still does zero work on the ions (option d).
2. Consequently no work is done by the magnetic force on the moving wire either (option c) - any kinetic energy the wire gains must come from an external agent (e.g. the battery), not from $\vec{B}$ itself. …

Why the magnetic force can never do work

For any charge $q$ moving with velocity $\vec{v}$ in a field $\vec{B}$:

The rate of work done is

because $\vec{v} \times \vec{B}$ is, by construction, perpendicular to $\vec{v}$. This is true for every moving charge, always - it is a geometric fact about the cross product, not a special property of wires.

Applying this to the wire

The free electrons inside the wire have a small drift velocity, but the crucial point is the ions of the lattice (positively charged, effectively fixed within the wire). If the wire as a whole is set moving by the external field, those ions now have a nonzero velocity too - the wire's velocity. Since the magnetic force is perpendicular to whatever velocity a charge actually has, it does zero work on these ions as well, even while the wire moves:

This is exactly statement (d): if the wire moves under the influence of $\vec{B}$, no work is done by the magnetic force on the ions, assumed fixed within the wire.

And since this holds for the motion of the wire as a whole (the ions carry it), it also confirms (c): if the wire moves under the influence of $\vec{B}$, no work is done by the force - i.e. any kinetic energy the wire gains must come from elsewhere (the battery/agent maintaining the current), never from the magnetic field itself.

Why (a) is false …

Method: "Magnetic Force Does No Work" -- Applying F dot v = 0 to Whatever Is Actually Moving

Any question built on the fact that magnetic forces do no work reduces to one check: identify the actual velocity of the charge in question -- even if that velocity is the bulk motion of an object the charge is embedded in -- and confirm the force is perpendicular to it.

Steps

Step 1: State the general no-work fact precisely

For any charge $q$ with velocity $\vec v$ in a field $\vec B$:

This is a geometric fact about the cross product ($\vec v\times\vec B$ is always $\perp\vec v$) -- it holds for every moving charge, unconditionally, regardless of what caused that velocity.

Step 2: Identify the velocity that actually matters

Don't assume "the charge" means only its microscopic drift velocity. If the charge is bound to a larger object (e.g. a lattice ion fixed within a wire) and that object itself starts moving, the charge's total velocity is now the object's velocity -- and Step 1 applies to that total velocity just as it would to a free charge.

Step 3: Separate "no work done" from "no effect at all" …