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A conducting rod of length 1m moves with a speed of 2m/s parallel to a straight long wire carrying 4A current. Axis of rod is kept perpendicular to the wire with its near end at a distance of 1m from the wire. Magnitude of induced emf in rod is $N \times 10^{- 7} \log 16$ volt. Find the value of N.

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answer is 4.

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Detailed Solution

Field at a distance x from wire is $B = \frac{μ_{0} l}{2 πx}$

and ${sϕ}_{B} = \int B . dA ∴ ϕ_{B} = \int_{1}^{2 l} \frac{μ_{0} lvt}{2 π} (\frac{dx}{x}) = \frac{μ_{0} lvt}{2 π} \log 2$

Hence, induced emf is $E = \frac{{dϕ}_{B}}{dt} = \frac{μ_{0} lv}{2 π} \log 2 = 16 \times 10^{- 7} \log 2 = 4 \times 4 \times 10^{- 7} \log 2$

So, $N = 4 = 4 \times 10^{- 7} \log 16$

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