Given,
i=i1+i2 .......(i)
∵ The length of part (ABC) is equal to length of part (ABD).
Here, resistance of part (ABC) is half that of part (ABD), i.e.
i1=2i2 ........(ii)
From eqs. (i) and (ii), we get
i1=,i2= Magnetic field due to AB and BC wire should be equal,
i.e.
BAB=BBC Magnetic field due to straight wire,
B=(sinα+sinβ) For wire AB,
From figure,
α=β=45°,and‌M‌O= ∴
BAB=.(sin45°+sin45°) ⇒
BAB=.=BBC Also, the magnetic field due to AD and DC wire should be equal, i.e.
BAD=BDC For wire AD,
BAD=.=BDC Net magnetic field at centre O is
Bnet=BAB+BBC−BAD−BDC Since,
(BAB+BBC)>(BAD+BDC) Bnet=.−. ⇒
Bnet= The direction of
Bnet will be inward to the plane of paper.