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Following figure shows a metal rod PQ re...

Following figure shows a metal rod PQ resting on the smooth ralls AH and positioned between the poles of a permanent magnet. The rails, the rod and the magnetic field are in three mutual perpendicular directions. A galvanometer G connects the rails through a switch K. Length of the rod 15 cm, B 0.50 T, resistance of the closed loop containing the rod `9.0 mOmega`. Assume the field to be uniform.
Suppose K is open and the rod is moved with a speed of `12"cm s"^(-1)` in the direction shown. Give the polarity and magnitude of the induced emf.

Text Solution

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Given, length of the rod `l=15cm=15xx10^(-2)m` Magnetic field `B=0.50T`
Resistance of the closed - loop containing the rod, `R=9mOmega=9xx10^(-3)Omega`
Velocity of rod `V=12" cm/s "12 xx 106(-2) m//s.`
The magnitude of the motional emf `e=BVl=0.50xx12xx10^(-2)xx15xx10^(-2),e=9xx10^(-3)V`
According to the Fleming's left hand rule, the direction of Lorentz force `(f =-e(V xx B))` on electrons in PQ is from P to Q. So, p would acquire positive charge and Q would acquire negative charge.
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