A wire of resistance `5.0 Omega` is used to wind a coil of radius 5 cm. The wire has a diameter 2.0 mm and the specific resistance of its material is `2.0xx10^(-7)Omega m`. Find the number of turns in the coil.

Explanation

Here, `R=5.0 Omega , r_(1) =5 xx10^(-2)m`,
`D=2.0xx10^(-3) m, rho =2.0xx10^(-7)Omega m`.
`R=rho l/(pi D^(2)//4)` or `l=(R pi D^(2))/(4 rho)`
Let n be the number of turns in the coil. Then total length of the wire used, `l= 2 pi r_(1) n`
or `n=l/2pi r_(1) = (R pi D^(2))/( 4 rho xx 2pi r_(1))=(R D^(2))/ (8 rho r_(1))`
`=(5.0 xx (2.0xx10^(- 3))^(2))/(8xx(2.0xx10^(-7)) xx(5xx10^(-2)))=250`