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Q.

A conducting resistance less rod is bent as parabola y=Kx2, where ‘K’ is a constant. A horizontal conductor of resistance per unit length λ starts sliding up on the parabola starting from x-axis with a constant acceleration a and the parabolic frame starting rotating with constant angular frequency ω about the axis of symmetry, both starting t = 0 as shown in the figure. There exist a uniform magnetic field B as shown in figure . The current induced in the rod when frame turns through an angle π4  is  Baπ(π12)n×122ωλ.  The value of ‘n’ is

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answer is 4.

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Detailed Solution

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Where the frame has turned through at angle θ, ϕ=BA  cosθ

where  A=0y2xdy=2K0yydy=43Ky3/2

since  y=12at2, Φ=B32Ka32t3cosθ

By faraday's law,Eind=dϕdt

When the frame turns through π4, I=EindR=π2(π12)B3/2K4ω192ω2K2aπλ=Baπ(π12)482ωλ

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