Question

Figure shows two parallel wires separated by a distance of $4.0cm$ and carrying equal currents of $10A$ along opposite directions. Find the magnitude of the magnetic field $B$ at the points $A_1,A_2,A_3$ and $A_4$.

Answer 1

Let the tow wires be positioned at $O$ & $P$

$R=OA,=\sqrt{(0.02{)}^2+(0.02{)}^2}=\sqrt{8\times {10}^{-4}}=2.828\times {10}^{-2}m$

a) $\overrightarrow{B}$ due to $Q,$ at $A_1=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times 0.02}=1\times {10}^{-4}T$ ($\perp r$ towards up the line)

$\overrightarrow{B}$ due to $P$, at $A_1=\frac{4\pi \times {10}^{-7}\times 10}{2\pi \times 0.06}=0.33\times {10}^{-4}T$ ($\perp r$ towards down the line)

net $\overrightarrow{B}=1\times {10}^{-4}-0.33\times {10}^{-4}=0.67\times {10}^{-4}T$

b) $\overrightarrow{B}$ due to $O$ at $A_2=\frac{2\times {10}^{-7}\times 10}{0.01}=2\times {10}^{-4}T$ $\perp r$ down the line

$\overrightarrow{B}$ due to $P$ at $A_2=\frac{2\times {10}^{-7}\times 10}{0.03}=0.67\times {10}^{-4}T$ $\perp r$ down the line

net $\overrightarrow{B}$ at $A_2=2\times {10}^{-4}+0.67\times {10}^{-4}=2.67\times {10}^{-4}T$

c) $\overrightarrow{B}$ at $A_3$ due to $O=1\times {10}^{-4}T$ $\perp r$ towards down the line

$\overrightarrow{B}$ at $A_3$ due to $P=1\times {10}^{-4}T$ $\perp r$ towards down the line

Net $\overrightarrow{B}$ at $A_3=2\times {10}^{-4}T$

d) $\overrightarrow{B}$ at $A_4$ due to $O=\frac{2\times {10}^{-7}\times 10}{2.828\times {10}^{-2}}=0.7\times {10}^{-4}T$ towards SE

$\overrightarrow{B}$ at $A_4$ due to $P=0.7\times {10}^{-4}T$ towards SW

Net $\overrightarrow{B}=\sqrt{(0.7\times {10}^{-4}{)}^2+(\mathrm{0..7}\times {10}^{-4}{)}^2}=0.989\times {10}^{-4}\approx 1\times {10}^{-4}T$