# Magnetic Field of Given Current Configuration

What is the magnetic field of the following current configurations:

a)

$B= (μ_0 I)/4π ∮ (dL ×r ̂)/r^2$

$B= (μ_0 I)/4π {∫_1^2…+∫_0^θ(a*dθ)/a^2+∫_3^4…+∫_θ^0(b*dθ)/b^2 }$

$B=(μ_0 I)/4π (0+1/a θ+0-1/b θ)$

$B=(μ_0 I)/4π θ(1/a-1/b)$

On 1->2 and 3->4 one has  $dL times hat r=0$.  On 2->3 and 4->1 one has $dL times hat r =dL=r*dθ$

b)

Definition (from wiki): line is aligned with a point if they are collinear so the brown angle in the figure is 45 degree. Magnetic field of the half loop is

$B(P)=1/2*(μ_0 I)/2R$    (standard known result for a loop)
Magnetic field of wire segment in fig 5.19 (example 5.5 from Griffiths)  (equation 5.37)

$B(P)=[(μ_0 I)/4πs]*[sin⁡ (θ_2)-sin (⁡θ_1)]$

For wire 1 one has

$B_1=[(μ_0 I)/4πR]*[sin ⁡90-0]=(μ_o I)/4πR$

For wire 2 one has

$B_2=[(μ_0 I)/4πd]*[sin 90-sin ⁡45]$   with $d=R*cos ⁡45=(R√2)/2$

$B_2=[(μ_0 I)/4πR]* 2/√2*(1-√2/2)=(μ_0 I)/4πR(2/√2-1)$

Altogether the 3 contributions (half loop and wires 1 and 2) one has (all 3 fields are into the page)

$B=(μ_0 I)/4R+(μ_0 I)/(2√2 πR)$