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Question

A rod PQ is connected to the capacitor plates. The rod is placed in a magnetic field (B) directed downward perpendicular to the plane of the paper. If the rod is pulled out of magnetic field with velocity V as shown in fig. Then, identify the correct statement :


A
Plate M will be positively charged
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B
Plate N will be positively charged
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C
both plates will be similarly charged
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D
No charge will be collected on either plates
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Solution

The correct option is A Plate M will be positively charged
The sign of Motional emf (ε) produced across the ends of a rod is given by:
V×B, so using right hand rule for cross product, the direction of thumb will point towards +ve terminal.
direction of V is ^i, direction of B is ^k
^i×^k=^j i.e along +ve y direction for (+ve) terminal of motional emf.
End P will be connected to +ve terminal of emf ε, hence plate M of capacitor will become positively charged.

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