Consider a wire carrying a steady current, I placed in a uniform magnetic field `vecB` perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do not work. This implies that

Options

A. Motion of charges inside the conductor is unaffected by B, since they do not aborb energy
B. some charges inside the wire move to the surface as a result of B
C. if the wire moves under the influence of B, no work is done by the force.
D. If the wire moves under the influence of B, no work is done by the magnetic force on the ions, assumed fixed within the wire.

Explanation

To solve the problem, we need to analyze the situation of a wire carrying a steady current placed in a uniform magnetic field that is perpendicular to its length. We will explore the implications of the magnetic forces acting on the charges within the wire and determine the correct statements regarding the work done by these forces.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a wire carrying a steady current \( I \).
- The magnetic field \( \vec{B} \) is uniform and perpendicular to the length of the wire.
- The current flows in one direction (let's say along the positive x-axis), while the magnetic field is directed in another (let's say along the positive z-axis).

2. **Identifying the Charges**:
- The wire consists of charged particles (typically electrons) that move in the opposite direction to the current. Therefore, if the current is in the positive x-direction, the electrons move in the negative x-direction.

3. **Calculating the Magnetic Force**:
- The magnetic force \( \vec{F} \) on the wire can be calculated using the formula:
\[ \vec{F} = I \vec{L} \times \vec{B} \]
- Here, \( \vec{L} \) is the length vector of the wire in the direction of the current. Since \( \vec{L} \) is along the x-axis and \( \vec{B} \) is along the z-axis, the angle between them is 90 degrees. Thus, the magnetic force will be directed along the negative y-axis.

4. **Analyzing the Work Done**:
- Magnetic forces do not do work on the charges because the force is always perpendicular to the displacement of the charges.
- The work done \( W \) by the magnetic force is given by:
\[ W = \vec{F} \cdot \vec{s} \]
- Since the force \( \vec{F} \) is perpendicular to the displacement \( \vec{s} \) of the charges, the work done is zero:
\[ W = F \cdot s \cdot \cos(90^\circ) = 0 \]

5. **Evaluating the Options**:
- **Option 1**: "Motion of charges inside the conductor is unaffected by \( B \) since they do not absorb energy."
- This is incorrect because the magnetic field does affect the motion of the charges, causing them to move in circular paths.
- **Option 2**: "Some charges inside the wire move to the surface as a result of \( B \)."
- This is correct. The magnetic force can cause some charges to redistribute towards the surface of the conductor.
- **Option 3**: "If the wire moves under the influence of \( B \), no work is done by the force."
- This is incorrect because if the wire moves, the magnetic force can do work on the wire.
- **Option 4**: "If the wire moves under the influence of \( B \), no work is done by the magnetic force on the ions."
- This is correct. The ions are fixed in the wire and do not move, so no work is done on them by the magnetic force.

### Conclusion:
The correct options are **Option 2** and **Option 4**.