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  • Find the current in the wire for the configuration shown in figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper.

Find the current in the wire for the configuration shown in figure. Wire PQ has negligible resistance. B, the magnetic field is coming out of the paper. \theta is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.

 

Answers (1)

Conducting electrons experience a magnetic force F_{m}=evB. So they move from P to Q within the rod. The end P of the rod becomes positively charged while end Q becomes negatively charged, hence an electric field is set up within the rod which opposes the further downward movement of electrons, i.e., an equilibrium F_{e}=F_{m}, i.e.,

eE=evB \; or\; E=vB\Rightarrow \text {Induced emf}e=El=Bvl\left [ E=\frac{V}{l} \right ]

If the rod is moved by making an angle \theta with the direction of the magnetic field or length. Induced emf,

e=Bvl\; \sin \theta

Emf induced across PQ due to its motion or change in magnetic flux linked with the loop change due to the change of enclosed area. The induced electric field E along the dotted line CD (perpendicular to both \vec{V}\; \text {and }\vec{B} and along \vec{V}\times \vec{B} ) =vB

Therefore, the motional emf along.

PQ=\left ( \text {length PQ} \right )\times \left ( \text {field along PQ} \right )

          = \left ( \text {length PQ} \right )\times \left ( \text {vB} \sin \theta \right )

          = \left ( \frac{d}{\sin \theta } \right )\times \left ( \text {vB} \sin \theta \right )=vBd

This induced emf make flow of current in closed circuit of resistnce R.

I=\frac{dvB}{R} and independent of \theta.

 

 

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infoexpert23

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