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 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.

Answer

A, B, D

Explanation

`m=pir^(2)Ldelta`...(i)
Here `L=2piR`
`epsilon=piR^(2)(dB)/(dt)=pi(L/(2pi))^(2)(dB)/(dt)=L^(2) /(4pi)(dB)/(dt)`
`i=epsilon/"Resistance"=(epsilon2piR)/(rho phi r^(2))`
After solving .
`i=m/(4pi rho delta)(dB)/(dt)`