Let the tow wires be positioned at O & P
R = OA,
= 2.828 × 10–2m
(a) vector B due to Q, at A1
(⊥r towards up the line)
vector B due to P, at A1
(⊥r towards down the line)
net vector B = 1 × 10–4 – 0.33 × 10–4 = 0.67 × 10–4T
(b) vector B due to O at A2
⊥r down the line
(b) vector B due to O at A2 = (2 x 10-7 x 10)/0.01 = 2 x 10-4T ⊥r down the line
net vector B at A2 = 2 × 10–4 + 0.67 × 10–4 = 2.67 × 10–4T
(c) vector B at A3 due to O = 1 × 10–4T ⊥r towards down the line
vector B at A3 due to P = 1 × 10–4 T ⊥r towards down the line
Net vector B at A3 = 2 × 10–4T
(d) vector B at A4 due to O
towards SE
vector B at A4 due to P = 0.7 × 10–4 T towards SW
= 0.989 ×10–4 ≈ 1 × 10–4 T
