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Please answer the question in image. This is a question from electromagnetic induction.

Shubham Kumar , 6 Years ago
Grade Select Grade
anser 1 Answers
Vikas TU
Dear student 
Speed = u
Magnetic field = B
Side = a
(a) The perpendicular component i.e. asinθ is to be taken which is ⊥r to velocity.
So, l = asinθ 30° = a/2.
Net ‘a’ charge = 4 x a/2 = 2a
So, induced emf = BϑI = 2auB
(b) Current = E/R = 2auB/R
Good Luck 
Last Activity: 6 Years ago
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