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Solution 1
The high level view can be given as below:
This can be further explained as below
Now we can derive and prove behaviour as
BA + BC=BB +BD = 2∗μ0 I /2 π
√ ( L
2 )2
+ ( L
2 )2
BA + BC=BB +BD =2∗μ0 I /2 π
√ L2
2

BA + BC=BB +BD = μ0 I
πL √ 1
2
BA + BC=BB +BD = μ0 I √2
πL
The direction is downwards
Solution 6
a)
Expression of electric field is
dE=k dQ
r2
Part 1: from (-r to 0)
Assuming θ is the angle with x axis then we can have
d Ex=dE cos ( θ )
d Ey=dE sin ( θ )

Integrating this we can have
Ex=∫
❑
❑
k dQ
r2 cos (θ )
Ex=∫
π
π /2
k dQ
r2 cos ( θ )
Ex=∫
π
π /2
k λr∗dθ
r2 cos ( θ )
Ex= λk
r
Similarly
E y=∫
❑
❑
k dQ
r2 sin ( θ )
E y=∫
π
π / 2
k dQ
r2 sin ( θ )
E y=∫
π
π / 2
k λr∗dθ
r2 sin ( θ )
E y=− λk
r
Now
λ= q
π
2 ∗r
λ= 2q
πr
Hence
Ex= λk
r =
2 q
πr k
r = 2qk
π r 2

E y=− λk
r =−2 qk
π r2
Part 2: from (0 to r)
Assuming θ is the angle with x axis then we can have
d Ex=dE cos ( θ )
d Ey=dE sin ( θ )
Integrating this we can have
Ex=∫
❑
❑
k dQ
r2 cos ( θ )
Ex=∫
π /2
0
k dQ
r2 cos ( θ )
Ex=∫
π /2
0
k λr∗dθ
r2 cos ( θ )
Ex=−λk
r
Similarly
E y=∫
❑
❑
k dQ
r2 sin ( θ )
E y=∫
π / 2
0
k dQ
r2 sin ( θ )
E y=∫
π / 2
0
k λr∗dθ
r2 sin ( θ )
E y=− λk
r
Now