A metallic square loop ABCD is moving in its own plane with velocity →V in a uniform magnetic field perpendicular to its plane as shown in fig. Then, an electric field induced
A
in AD, but not in BC
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B
in AB and CD both.
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C
neither in AD nor in BC
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D
in AD and BC both.
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Solution
The correct option is D in AD and BC both. ⇒The motional emf induced across the ends of moving rod is : ε=VBl⊥ l⊥=length of rod perpendicular to its velocity. In this case the length of rods AB and DC is parallel to direction of velocity, ∴l⊥=0 for AB and DC. ⇒ε=0 for AB and DC However for rod AD and BC the l⊥ exists, hence emf (ε) is induced in these segments.
∴ Induced Electric field will be present in AD and BC ,since electric field is associated with potential difference (emf).