Question
Consider a wire carrying a steady current, I placed in a uniform magnetic field B perpendicular to its length. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that:
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Answer
- Some charges inside the wire move to the surface as a result of B.
- 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.
Magnetic force on a conductor of length l carrying a current I placed in a uniform magnetic field B is given by, $\text{F}=\text{I}(\text{l}\times\text{B})\text{ or }|\text{F}|=\text{I }|\text{l}| \ \ |\text{B}|\sin\theta$.
The direction of force is given by Fleming's left hand rule & F is perpendicular to the direction of magnetic field B, Therefore, work done by the magnetic force on the ions is zero.
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