A copper wire having a resistance of `0*02Omega` per metre is used to wind a 500 turns solenoid of radius `2*0cm` and length `30cm`. What should be the `emf` of the battery which when connected across the solenoid would produce a magnetic field at `10^-2T`, near the centre of the solenoid.

Answer

`6V`

Explanation

Length of wire used, `L=2pirxx` no. of turns
`=2pixx(2xx10^-2)xx500m`
Resistance per unit length `=0*02Omegam^-1`
Total resistance of wire,
`R=2pixx(2xx10^-2)xx500xx*02=0*4piOmega`
No. of turns per unit length,
`n=(500)/(30xx10^-2)=(5000)/(3)m^-1`
As `B=mu_0nI=mu_0(n epsilon)/(R)`
So `epsilon=(BR)/(mu_0n)=(10^-2xx(0*4pi))/((4pixx10^-7)xx(5000//3))=6V`