InterviewSolution
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InterviewSolution
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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 not work. This implies thatA. motion of charges inside the conductor is unaffected by `vecB` since they do not absorbB. Some charges inside the wire move to the surface as a result of `vecB`C. if the wire moves under the influence of `vecB`, no work is done by the forceD. 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 |
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Answer» Correct Answer - B::D (a) Motion of charges inside the conductor is affected by magnetic field `vecB`, due to magnetic force `vecF`, given by `vecF=q(vecvxxvecB)` (b) Due to magnetic force, some charges inside the wire move to the surface of wire. (c) The force on wire of length l, carrying current I when subjected to magnetic field `vecB` is, `vecF=I(veclxxvecB)`. It acts perpendicular to the plane containing `vecl` and `vecB` and is directed as given by Right Hand rule. If the wire moves under the influence of `vecB` at an angle `theta`, where `theta=90^@`, then work done, `W=Fscostheta`, can not be zero. (d) When wire moves under the influence of `vecB`, then displancement of the ions is perpendicular to the magnetic force `vecF`. Therefore work done is zero. |
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