#16
Using Ampere's Law, we have,
2 pi.y.B = o
Io sinwt
=> Magnetic field B = (o
Io/ 2pi.y) sinwt
where y is the perpendicular distance from the power line to any
point.
Therefore, the magnetic flux collected by the wire loop
is:
= (o.L /
2pi) ln(y2/y1) Io sin(wt)
where y1 = 5 and y2 = 7 as per given.
Therefore, using Faraday's Law,
Voltage, V = (oL / 2pi)
ln(y2/y1) w Io cos(wt)
Solving for L we get
L = ( 2pi.Vo /oln(y2/y1)w.Io)
=> L =
2pi*170/4pi*10-7ln(7/5).2pi*60*55*103 =
59.26m
#16 A 22.0-cm-diameter coil consists of 28 turns of circular copper wire 2.6 mm in diameter....
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