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A square loop of side ‘a’ is placed in the same plane as a long straight wire carrying a current ‘i'. The center of the loop is at a distance ‘r’ from the wire where $r > > a .$ The loop is moved away from the wire with a constant velocity ‘v’. The induced emf in the loop is $\frac{{kμ}_{0} {ia}^{2} v}{2 {πr}^{2}} .$ Then the value of ‘ $k$ ’ is equal to ………..

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Detailed Solution

Magnetic flux likes with the loop is $ϕ = B . S = \frac{μ_{0}}{2 π} \frac{i}{r} (a^{2}) = \frac{μ_{0} i a^{2}}{2 π r}$

$∴ e = | \frac{d ϕ}{d t} | = \frac{μ_{0} i a^{2}}{2 π r^{2}} \frac{d r}{d t} = \frac{μ_{0} i a^{2}}{2 π r^{2}} v (\frac{d r}{d t} = v)$

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