A small square loop of wire of side is placed inside a large square loop of wire. L(L >> ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:
A small square loop of wire of side is placed inside a large square loop of wire. L(L >> ). Both loops are coplanar and their centres coincide at point O as shown in figure. The mutual inductance of the system is:
Assuming current 'i' In outer loop magnetic field at centre
M =
=
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Kindly go through the solution
Bv = B sin 60°
->
M = φ? /I? = (B? A? )/I? = [ (μ? I? /2R? )πR? ²]/I?
[Diagram of two concentric coils]
M = (μ? πR? ²)/ (2R? )
M ∝ R? ²/R?
(A) The magnet's entry
R =
L = 2 mH
E = 9V
Just after the switch ‘S’ is closed, the inductor acts as open circuit.
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Physics Ncert Solutions Class 12th 2023
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