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. `(4pi m)/(rho sigma)(dB)/(dt)`
B. `(m)/(4 pi rho sigma)(dB)/(dt)`
C. `(m)/(rho sigma)(dB)/(dt)`
D. `(2 pi m)/( rho sigma)(dB)/(dt)`

Answer

B

Explanation

Let `r` is the radius of the loop. The magnetic flux, at a time `t`, through the loop is given by
`phi = B(pi r^(2)) , e = (d phi)/(dt) = pi r^(2)((dB)/(dt))`
(the numerical value of induced emf)
Induced current in the loop is given by
`i = (e)/(R ) = (pi r^(2))/(R )(dB)/(dt)` ......(i)
where `R` is the resistance of the wire let a is the cross-section area and the length of the wire. Then
`R = rho(l)/(pi a^(2))` ........(ii)
where `l = 2 pi r , m = pi a^(2)l sigma`
On putting the value of `a^(2)` in eq. (ii), we get
`i = (m)/(4 pi rho sigma)(dB)/(dt)`