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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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some charges inside the wire move to the surface as a result of $\mathbf{B}$
,
if the wire moves under the influence of $\mathbf{B}$, no work is done by the magnetic force on the ions, assumed fixed within the wire.
some charges inside the wire move to the surface as a result of $\mathbf{B}$
,
if the wire moves under the influence of $\mathbf{B}$, no work is done by the magnetic force on the ions, assumed fixed within the wire.
Magnetic forces on a current carrying conductor, consider wire carrying a steady current I, placed in a uniform magnetic field $\mathrm{B}$, perpendicular to its length isF=
IlBsing $0^{\circ}=\mathrm{IlB}$
The direction of force is get by Fleming's left hand rule and $\mathrm{F}$ is perpendicular to the direction of magnetic field $\mathrm{B}$. Hence, work done by the magnetic force on the ions is 0 .
IlBsing $0^{\circ}=\mathrm{IlB}$
The direction of force is get by Fleming's left hand rule and $\mathrm{F}$ is perpendicular to the direction of magnetic field $\mathrm{B}$. Hence, work done by the magnetic force on the ions is 0 .
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