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 no work. This implies that,

Consider a wire carrying a steady current, I placed in a uniform magne

1.	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 no work. This implies that,
Answer» motion of charges inside the CONDUCTOR is unaffected by `vecB` since they do not absorb energy. some charges inside the wire move to the SURFACE as a result of `vecB`. if the wire moves under the INFLUENCE of `vecB`, no work is done by the FORCE. if the wire moves under the influence of `vecB`, no work is done by the magnetic force on the ions, assumed fixed within the wire. Solution :CURRENT carrying wire (I) is held in magnetic field as it is perpendicular to magnetic field. So magnetic force act on it is, `F=IlBsintheta` `theta=90^(@)" "(becauseIvecl_\|_vecB)` `thereforeF=IlB` According to Fleming.s left hand rule the direction of magnetic force is perpendicular to magnetic field. So, work done by magnetic force on ions are zero. So, options (B), (D) are true.

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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 no work. This implies that,

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