A magnetic field induction is changing in magnitude in a region at a constant rate `dB//dt`. A given mass `m` of copper drawn into a wire and formed into a loop is placed perpendicular to the field. If the values of specific resistance and density of copper are `rho` and `sigma` respectively, then the current in the loop is given by :

Options

A. (a) `(m)/(2pirhod)(dB)/(dt)`
B. (b) `(m)/(4pia^(2)r)(dB)/(dt)`
C. (c) `(m)/(4piad)(dB)/(dt)`
D. (d) `(m)/(4pirhod)(dB)/(dt)`

Answer

D

Explanation

(d) `xi = (dphi)/(dt) `rarr` I = (xi)/(R )(d)/(dA)(BA) = (A)/(R )(dB)/(dt)`
where `pir^(2)` = area of the loop of radius `r` and `R` = resistance of the loop of length `(2pir)` and area of cross section `pia^(2)`.
`R = (rhol)/(pia^(2)) = (rho(2pir))/(pia^(2))`
Further mass of wire is `m = (pia^(2))(2pir)(d)`
`i = ((pia^(2))(pir^(2)))/(rho(2pir))(dB)/(dt)`
`i = ((pia^(2))(pir^(2)))/(4pirho)(dB)/(dt) rarr i = (m)/(4pirhod)(dB)/(dt)`