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Consider a wire carrying a steady curren...

Consider a wire carrying a steady current I placed in a uniform magnetic field `vecB`. Consider the charges inside the wire. It is known that magnetic forces do no work. This implies that

A

motion of charges inside the wire move the surface as a result of `vecB`.

B

some charges inside the wire move the surface as a result of `vecB`.

C

if the wire moves under the influence of `vecB`,no work is done by the force

D

if the wire moves under the influence of `vecB`, no worik is done by the magnetic force on the ions, assumed fixed within the wire.

Text Solution

Verified by Experts

The correct Answer is:
B, D
`vecF-q(vecvxxvecB)`
Also `vecF=I(veclxx vecB)`
Direction of force due to magnetic field is perpendicular to the displacement of the charge.
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