1i think it is in h.c verma ( EMI )
Becoz the words are a bit unclear.."Metallic square of edge l in a vertical plane. Uniform Magnetic field perpendicular to plane of figure. The sqaure is being deformed into a rhombus due to force exerted on it by the 2 boys. Corners being pulled at uniform speed u. Find induced Emf when angle of these corners is reduced to 60."
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5 Answers
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(In case the diagram is too unclear, I have basically just defined θ, Lsinθ and Lcosθ)
φ=∫B.dA
dφ=B.(4* (.5* lcosθ* lsinθ))
=2Bl2cosθsinθ
= Bl2sin2θ
Now,
dx/dt = u = -lsinθ * dθ/dt
dφ/dt = -Bl2* 2 cos2θ * dθ/dt
Subsituting Value of dθ/dt,
dφ/dt = -2Blucos2θ/sinθ
Now, when angle between corners is 60°, θ=60/2=30
Subsituting value for θ, and taking absolute value, EMF= 2Blu
Absolute Value can be taken as polarity of EMF does not matter.