An infinite carrying conductor is bent into three segment (1), (2) and (3) as shown in Fig. If it carries a current i, find the magnetic induction at the origin.

Explanation

Net magnetic field at O will be the vector sum of magnetic
fields due to wires (1) , (2) and (3).
Magnetic field due to wire (1),
`vecB_1=(mu_0)/(4pi) I/a[sin 90^@-sin45^@](-hatj)`
`=(mu_0)/(4pi) i/a[1-1/sqrt2] (-hatj)`
Magnetic field due to wire (2),
`vecB_2=(mu_0)/(4pi) I/a[sin 45^@-sin0^@](hatj)`
`=(mu_0)/(4pi) i/a[1/sqrt2] (hatj)`
Magnetic field due to wire (3),
`vecB_3=(mu_0)/(4pi) I/a (hatk)`
Hence, net magnetic field `vecB=vecB_1+vecB_2+vecB_3`
`implies (mu_0)/(4pi) i/a[(sqrt2-1)hatj+hatk]`