Two long wires are at a distance 'd' apart carries equal and antiparallel current 'i'. Calculate the magnetic field induction at point 'p' through distance 'R'?
Solution
∣ B 1 → ∣ = ∣ B 2 → ∣ = μ 0 i 2 π l l , where l = ( d 2 ) 2 + R 2 . \left| \overrightarrow {B _ {1}} \right| = \left| \overrightarrow {B _ {2}} \right| = \frac {\mu_ {0} i}{2 \pi} \frac {l}{l}, \text{where } l = \sqrt {\left(\frac {d}{2}\right) ^ {2} + R ^ {2}}. ∣ ∣ B 1 ∣ ∣ = ∣ ∣ B 2 ∣ ∣ = 2 π μ 0 i l l , where l = ( 2 d ) 2 + R 2 .
If the coordinates x x x y y y
∣ B 1 y ∣ = ∣ B 2 y ∣ from symmetric configuration \left| B _ {1 y} \right| = \left| B _ {2 y} \right| \quad \text{from symmetric configuration} ∣ B 1 y ∣ = ∣ B 2 y ∣ from symmetric configuration ∣ B x ∣ = ∣ B 2 x ∣ \left| B _ {x} \right| = \left| B _ {2 x} \right| ∣ B x ∣ = ∣ B 2 x ∣
Therefore { B y = 0 B x = − ∣ B 1 → ∣ cos θ − ∣ B 2 → ∣ cos θ = − 2 ∣ B 1 → ∣ cos θ , \begin{cases} B_y = 0 \\ B_x = -\left|\overrightarrow{B_1}\right| \cos \theta - \left|\overrightarrow{B_2}\right| \cos \theta = -2\left|\overrightarrow{B_1}\right| \cos \theta, \end{cases} { B y = 0 B x = − ∣ ∣ B 1 ∣ ∣ cos θ − ∣ ∣ B 2 ∣ ∣ cos θ = − 2 ∣ ∣ B 1 ∣ ∣ cos θ ,
Where B → = B 1 → + B 2 → , cos θ = d 2 l \overrightarrow{B} = \overrightarrow{B_1} +\overrightarrow{B_2},\cos \theta = \frac{d}{2l} B = B 1 + B 2 , cos θ = 2 l d
∣ B → ∣ = ∣ B x ∣ = 2 x ( μ 0 i l 2 π l ) ∗ d 2 l = μ 0 i 2 π d l 2 = μ 0 i 2 π d ( d 2 ) 2 + R 2 = μ 0 i 2 π 4 d d 2 + 4 R 2 = 2 μ 0 i d π ( 4 R 2 + d 2 ) . \left| \overrightarrow {B} \right| = \left| B _ {x} \right| = 2 x \left(\frac {\mu_ {0} i l}{2 \pi l}\right) * \frac {d}{2 l} = \frac {\mu_ {0} i}{2 \pi} \frac {d}{l ^ {2}} = \frac {\mu_ {0} i}{2 \pi} \frac {d}{\left(\frac {d}{2}\right) ^ {2} + R ^ {2}} = \frac {\mu_ {0} i}{2 \pi} \frac {4 d}{d ^ {2} + 4 R ^ {2}} = \frac {2 \mu_ {0} i d}{\pi (4 R ^ {2} + d ^ {2})}. ∣ ∣ B ∣ ∣ = ∣ B x ∣ = 2 x ( 2 π l μ 0 i l ) ∗ 2 l d = 2 π μ 0 i l 2 d = 2 π μ 0 i ( 2 d ) 2 + R 2 d = 2 π μ 0 i d 2 + 4 R 2 4 d = π ( 4 R 2 + d 2 ) 2 μ 0 i d . 
