Question

Consider a wire carrying a steady current 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

• A. motion of charges inside the conductor is unaffected by B since they do not absorb 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. it the wire moves under the influence of B no work is done by the magnetic forces on the ions assumed fixed the wire

Answer 1

Answer: (B), (D)

some-charges inside the wire move to the surface as a result of B
it the wire moves under the influence of B no work is done by the magnetic forces on the ions assumed fixed the wire

Due to the magnetic force, $\overrightarrow{F_m}=q(\overrightarrow{v}\times \overrightarrow{B}),$ some charges move to the surface.

Also, as $\overrightarrow{F_m}$ is always perpendicular to $\overrightarrow{B}$ thus work done by magnetic force only is always zero.