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A long straight wire is parallel to one edge as shown in fig. If the current in the long wire varies with time as $I = I_{0} e^{- t/ τ}$ , what will be the induced emf in the loop

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\frac{μ_{0} b I}{π τ} l n (\frac{d + a}{d})

\frac{μ_{0} b I}{2 π τ} l n (\frac{d + a}{d})

\frac{μ_{0} b I}{π τ} l n (\frac{d}{d + a})

\frac{2 μ_{0} b I}{π τ} l n (\frac{d + a}{d})

answer is B.

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Detailed Solution

Magnetic field in the strip ‘dx’ due to the current

i = i_{0} e^{- t/ λ} is B = \frac{μ_{0} i}{2 π x}

Flux $ϕ = BA$

$\Rightarrow ϕ = \frac{μ_{0} i}{2 π x} bd x$

$\Rightarrow ϕ = \frac{μ_{0} i b}{2 π} \int_{d}^{(d + a)} \frac{1}{x} d x = \frac{μ_{0}}{2 π} i b l n [\frac{d + a}{d}]$

Induced emf $ε = \frac{- d ϕ}{dt} \Rightarrow ε = \frac{- μ_{0}}{2 π} i_{0} b$

$l n (\frac{d + a}{d}) \frac{d}{d t} (e^{- t/ τ})$

$ε = \frac{μ_{0} I b}{2 π τ} l n (\frac{d + a}{d})$

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