A long, circular pipe, with an outside radius R, carries a (uniformly distributed) current `i_0` (into the paper as shown in Fig.) A wire runs parallel to the pipe at a distance 3R from centre to centre. Calculate the magnitude and direction of the current in the wire that would cause the resultant magnetic field at point P to have the same magnitude, but the opposite direction, as the resultant field at the centre of the pipe.

Explanation

Suppose wire carries current `i_1` and in the same direction as that
of the pipe.
Magnetic field at the centre of the pipe=`-(mu_0i_1)/(6piR)hatj=(mu_0i_1)/(6piR)(-hatj)`

Megnetic field at point P
`(mu_0i_0)/(2pi(2R))hatj-(mu_0i_1)/(2piR)hatj=((mu_0i_0)/(4piR)-(mu_0i_1)/(2piR))hatj`
Now magnitudes are equal
`(mu_0i_1)/(6piR)=(mu_0i_0)/(4piR)-(mu_0i_1)/(2piR) implies i_1=(3i_0)/8`