A long thin wire carrying a varying current `I = i_(0) sin omegat` lies at a distance `y` above one edge of a rectangular wire loop of length `L and W` lying in the `x-z` plane. What emf is induced in the loop?

Answer

`(mu_(0))/(4pi)i_(0)W omega cos omega t1n((L^(2) + y^(2))/(y^(2)))`

Explanation

The magnetic field at point `P` [Fig. S3.27(b)] is
`B = (mu_(0))/(2pi)(i)/sqrt(y^(2) + z^(2))`
The magnetic flux through the shaded strip in Fig. S3.27(a) is
`dphi = (W dz)(mu_(0))/(2pi)(i)/sqrt(y^(2) + z^(2))sin theta`
where `sin theta = (z)/sqrt(y^(2) + z^(2))`
`:.` total magnetic flux through rectangular loop is
`phi = int_(0)^(L)(mu_(0))/(2pi)(i_(0)sin omegat Wzdz)/(y^(2) + z^(2))`
`= (mu_(0))/(4pi)((y^(2) + L^(2))/(y^(2)))i_(0)sin omegat`
(a)
`:.` induced emf in the loop is
`e = (dphi)/(dt) = (mu_(0))/(4pi)i_(0)W omegat 1n((L^(2) + y^(2))/(y^(2)))`
Alternate method to find flux `phi`:
Flux through the given loop will be same as the flux through a cylindrical region from radius `y` to radius `sqrt(L^(2) + y^(2))`

So, `phi = "in"t (mu_(0)i)/(2pix)W dx` `rarr` ` phi = phi = (mu_(0)i)/(2pi)W int _(y)^(sqrt(L^(2) + y^(2)) (dx)/(x)`
`rarr` ` phi = (mu_(0)(i_(0)sin omegat)W)/(2pi)log_(e)[sqrt(L^(2) + y^(2))/(y)]`
`= (mu_(0)W(i_(0)sin omegat))/(4pi)log_(e)[(L^(2) + y^(2))/(y)]`
Now induced emf can be found .