Find the magnetic induction at the point `O` if the wire carrying a current `I = 8.0 A` has the shwon in Fig.



The radius of the curved part of the wire is `R = 100 mm`, the linear parts of the wire are very long.

Explanation

(a) `vec(B_(0)) = vec(B_(1)) + vec(B_(2)) + vec(B_(3))`
`= (mu_(0))/(4pi) (i)/(R) (-vec(k)) + (mu_(0))/(4pi) (i)/(R) pi (-vec(i)) + (mu_(0))/(4pi) (i)/(R) (-vec(k))`
`= -(mu_(0))/(4pi) (i)/(R) [2vec(k) + pi l]`
So, `|vec(B_(0))| = (mu_(0))/(4pi) (i)/(R) sqrt(pi^(2) + 4) = 0.30 mu T`
(b) `vec(B_(0)) = vec(B_(1)) + vec(B_(2)) + vec(B_(3))`
`= (mu_(0))/(4pi) (i)/(R) (-vec(k)) + (mu_(0))/(4pi) (i)/(R) pi (-vec(i)) + (mu_(0))/(4pi) (i)/(R) (-vec(i))`
`= -(mu_(0))/(4pi) (i)/(R) [vec(k) + (pi + 1) vec(i)]`
So,
`|vec(B_(0))| = (mu_(0))/(4pi) (i)/(R) sqrt(1 + (pi + 1)^(2)) = 0.34 mu T`
(c) Here using the law of parallel resistances
`i_(1) + i_(2) = 1` and `(i_(1))/(i_(2)) = (1)/(3)`,
So, `(i_(1) + i_(2))/(i_(2)) = (4)/(3)`
Hence `i_(2) = (3)/(4) i`, and `i_(1) = (1)/(4) i`
Thus `vec(B_(0)) = (mu_(0))/(4pi) (i)/(R) (-vec(k)) + (mu_(0))/(4pi) (i)/(R) (-vec(j)) + [(mu_(0))/(4pi) ((3pi)/(2)) (i_(1))/(R) (-vec(i)) + (mu_(0))/(4pi) ((pi//2) i_(2))/(R) vec(i)]`
`= (-mu_(0))/(4pi) (i)/(R) (vec(j) + vec(k)) + 0`
Thus, `|vec(B_(0))| = (mu_(0))/(4pi) (sqrta(2)i)/(R) = 0.11 mu T`