
(a) For each of the wire
Magnitude of magnetic field

For AB ⊙ for BC⊙ For CD ⊗ and for DA ⊗.
The two ⊙ and 2⊗ fields cancel each other. Thus Bnet = 0
(b) At point Q1
due to (1)B

= 4 × 10–5⊙
due to (2)B

= (4/3) × 10–5⊙
due to (3)B

= (4/3) × 10–5⊙
due to (4) B =

= 4 × 10–5⊙
Bnet = [4 + 4 + (4/3) + (4/3)] × 10–5 = 32/3 × 10–5 = 10.6 × 10–5 ≈ 1.1 × 10–4T
At point Q2

At point Q3

Bnet = [4 + 4 + (4/3) + (4/3)] × 10–5 = 32/3 × 10–5 = 10.6 × 10–5 ≈ 1.1 × 10–4T
