Magnetic field due to a straight current carrying conductor of finite length
B=‌‌(sin‌θ1+sin‌θ2)....(i)
(i) Magnetic field due to conductor
DA.
Here,
d=‌=2m θ1=θ2=tan−1(‌)=54.73∘ ∴sin‌θ1=sin‌θ2=sin‌54.73∘=0.816
and
‌‌I=5A (given)
From Eq. (i),
B1=‌×‌(0.816+0.816) B1=4.08×10−7T Similarly, magnetic field due to conductor
BC B2=4.08×10−7T (ii) Magnetic field due to conductor
AB.
Here,
‌‌d=‌=2√2m θ1=θ2=tan−1(‌)=35.26∘ ∴sin‌θ1=sin‌θ2=sin‌35.26∘=0.577
and
I=5A (given)
From Eq. (i),
B3=‌×‌(0.57+0.57) B3=2.04×10−7T Similarly magnetic field due to conductor
CD B4=2.04×10−7T Hence, magnitude of induction field vector
B at the intersection of the diagonals
B=B1+B2+B3+B4 B=(4.08+4.08+2.04+2.04)×10−7
B=12.24×10−7T or
B=1.2×10−6T