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A conducting resistance less rod is bent as parabola $y = K x^{2},$ 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 $\frac{π}{4} i s \frac{B a π (π - 12)}{n \times 12 \sqrt{2} ω λ} .$ The value of ‘ $n$ ’ is

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answer is 4.

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Detailed Solution

Where the frame has turned through at angle $θ, ϕ = B A \cos θ$

$w h e r e A = \int_{0}^{y} 2 x d y = \frac{2}{\sqrt{K}} \int_{0}^{y} \sqrt{y} d y = \frac{4}{3 \sqrt{K}} y^{3 / 2}$

$\sin c e y = \frac{1}{2} a t^{2}, ∴ Φ = \frac{B}{3} \sqrt{\frac{2}{K}} a^{\frac{3}{2}} t^{3} \cos θ$

By faraday's law, $E_{i n d} = \frac{d ϕ}{d t}$

When the frame turns through $\frac{π}{4}, I = \frac{E_{i n d}}{R} = \frac{π^{2} (π - 12) B^{3 / 2} \sqrt{K} 4 ω}{192 ω^{2} \sqrt{K} \sqrt{2 a} π λ} = \frac{B a π (π - 12)}{48 \sqrt{2} ω λ}$

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