A magnetic field induction is changing in magnitude at a constant rate `dB//dt` . A given mass `m` of copper is drawn into a wire into a wire of radius `alpha` and formed into a loop of radius `r` is placed perpendicular to the field. Show that induced current in the loop is given by `i(m)/(4pipdelta)(dB)/(dt)`
`p` : resistivity, `delta` : density of copper.

Explanation

`m=deltapialpha^(2)l=deltapialpha^(2).2pir`
`alpha` : radius of cross-section of wire
`phi=BA=Bpir^(2)`
`e=(dphi)/(dt)=pir^(2)(dB)/(dt)`
Resistance of loop `R=(p(2pir))/(pialpha^(2))=(2pr)/(alpha^(2))`
Induced current
`i=(e)/(R)=(pir^(2)(dB)/(dt))/(2pr//alpha^(2))`
`=(piralpha^(2)(dB)/(dt))/(2P)`
From (i), `piralpha^(2)(m)/(2pidelta)`
`i=(m)/(4pipdelta).(dB)/(dt)`