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A conducting rod length 1m moves with a ...

A conducting rod length `1m` moves with a speed of `2m//s` parallel to a straight long wire carrying `4 A` 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 xx 10^(-7) log 16 "volt"`. Find the value of `N`.

Text Solution

Verified by Experts

The correct Answer is:
D
Field at distance `x` from the wire is `B = (mu_(0)i)/(2pi x)`
and `phi_(0) = int B.dA`
`:. phi_(B) = int_(l)^(2l)(mu_(0)i vt)/(2pi) ((dx)/x)=(mu_(0)i vt)/(2pi)"log"2`
Hence induced emf is
`E = (d phi_(B))/(dt) = (mu_(0)i v)/(2pi)"log" 2 = 16 xx 10^(-7) log2`
`:. N = 4`
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