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. `(4pim)/(rhosigma)(dB)/(dt)`
B. `m/(4pirhosigma)(dB)/(dt)`
C. `m/(rhosigma)(dB)/(dt)`
D. `(2pim)/(rhosigma)(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(pir^2),e=(dphi)/(dt)=pit^2((dB)/(dt))`
the numerical valu of induced emf
Induced current in the loop is given by
`i=e/R=(pir^2)/R(dB)/(dt)` .......(i)
where R is the resistance of the wire let a is the cross -section area and 1 the length of the wire. Then
`R=rhol/(pia^2)`........(ii)
where `l=2pir,m=pia^2lsigma`
on puttting the value of `a^2`in eq (ii) we get
`i=m/(4pirhosigma)(dB)/(dt)`