NEET Class neet Answered
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Asked by anjanakurup728 | 21 Dec, 2019, 09:36: PM
Expert Answer
Reduction factor of tangenet galvanometer K = ( 2 a Bh ) / (μo n )
where a is radius of coil, Bh is horizontal component of earth's magnetic field,
μo is magnetic permeability of free space and n is number of turns.
If KA and KB are reduction factors ofgiven two tangent galvanometers, then we have
KA : KB = ( 2 aA Bh ) / (μo nA ) : ( 2 aB Bh ) / (μo nB ) = (aA/ aB ) / (nB / nA ) = (1/3)(2/1) = 2/3
Hence KB > KA
Current I in tangent galvanometer is given by I = K tanθ
IA : IB = KA tanθA : KB tanθB
since the two tangenet galvanometers are connected in series, current in each galvanometer is same.
hence, tanθA : tanθB = KB : KA = 3/2 , i.e., θA > θB
Answered by Thiyagarajan K | 22 Dec, 2019, 09:56: AM
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