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Question: A rectangular loop of length l and breadth b is placed at a distance of x from infinitely long wire ...

A rectangular loop of length l and breadth b is placed at a distance of x from infinitely long wire carrying current i such that the direction of current is parallel to breadth. If the loop moves away from the current wire in a direction perpendicular to it with a velocity n, the magnitude of the emf in the loop is (µ0 = permeability of free space)

A

μ0iν2πx(l+bb)\frac{\mu_{0}i\nu}{2\pi x}\left( \frac{\mathcal{l +}b}{b} \right)

B

μ0i2ν4π2xlog(bl)\frac{\mu_{0}i^{2}\nu}{4\pi^{2}x}\log\left( \frac{b}{\mathcal{l}} \right)

C

μ0ilbν2πx(l+x)\frac{\mu_{0}i\mathcal{l}b\nu}{2\pi x\mathcal{(l +}x)}

D

μ0ilbν2πlog(x+lx)\frac{\mu_{0}i\mathcal{l}b\nu}{2\pi}\log\left( \frac{x + \mathcal{l}}{x} \right)

Answer

μ0ilbν2πx(l+x)\frac{\mu_{0}i\mathcal{l}b\nu}{2\pi x\mathcal{(l +}x)}

Explanation

Solution

Flux with small coil df = (µ0I2πr)\left( \frac{µ_{0}I}{2\pi r} \right) (bdr)

\ Flux with complete coil f = xx+ldφ\int_{x}^{x + \mathcal{l}}{d\varphi}

= µ0Ib2π\frac{µ_{0}Ib}{2\pi}log (x+lx)\left( \frac{x + \mathcal{l}}{x} \right)

e = – dφdt\frac{d\varphi}{dt}= –µ0Ib2π\frac{µ_{0}Ib}{2\pi}.(0lx2dxdt)\left( 0 - \frac{\mathcal{l}}{x^{2}}\frac{dx}{dt} \right)

= µ0Iblv2πx(x+l)\frac{µ_{0}Ib\mathcal{l}v}{2\pi x(x + \mathcal{l})}