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Question
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. θ is a fixed angle made by PQ travelling smoothly over two conducting parallel wires separated by a distance d.
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Solution
This problem is based upon the motional emf. Consider a conducting rod of length l moving with a uniform velocity v perpendicular to a uniform magnetic field B bar, directed into the plane of the paper. Let the rod be moving to the right as shown in figure. The conducting electrons also move to the right as they are trapped within the rod.
Conducting electrons experience a magnetic force Fm = 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 is reached and in equilibrium Fe = Fm, i.e., eE = evB or R = vB ⇒ Inducted emf e = `El = Bwl [E = V/l]`
If rod is moved by making an angle θ with the direction of magnetic field or length. Induced emf, `e = Bvl sin θ`
 (A) |
 (B) |
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 `vecv` and `vecB` along `vecv xx vecB`) = `vB`
Therefore, the emotional emf along
PQ = (length PQ) × (field along PQ)
= (length PQ) × (vB sin θ)
= `(d/sin theta) xx (vB sin theta) = vBd`
This induced emf make flow of current in closed circuit of resistance R.
`I = (dvB)/R` and is independent of θ.
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