91Ó°ÊÓ

• Current$i=10\text{A}$
• Distance $l=0.50\text{m}$

We use the formula of magnetic field due to an infinitely long wire at perpendicular distance to calculate the magnetic field due to the wire. Using this formula we can also find the current in the wire.

Formulae:

$\mathit{B}\mathbf{=}\frac{{\mathbf{μ}}_{\mathbf{0}}\mathbf{i}}{\mathbf{2}\mathbf{π}\mathbf{d}}$

Using the right-hand rule, the magnetic field ${\stackrel{→}{B}}_1$ produced by wire 1(the wire at the bottom left) at role="math" localid="1662998840239" $a$is at an angle $\mathrm{ϕ}=150°$measured clockwise from $+x$axis, and the field produced by wire 2( the wire at the bottom right) is at the angle $\mathrm{ϕ}=210°.$

By symmetry role="math" localid="1662999221301" $(\stackrel{→}{B_1}=\stackrel{→}{B_2})$ we observe that only role="math" localid="1662999206462" $x-$component survives, giving the net magnetic field.

role="math" localid="1662999882956" $$\begin{gathered} \stackrel{→}{B}={\stackrel{→}{B}}_1+{\stackrel{→}{B}}_2 \\ \stackrel{→}{B}=\left(2\frac{{\mathrm{μ}}_{0}i}{2\mathrm{π}l}\text{cos}150°\right)\hat{\mathrm{i}} \\ \stackrel{→}{\mathrm{B}}=\left(\frac{\left(1.26×{10}^6{\displaystyle \raisebox{1ex}{$\mathrm{T}.\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}\right)\left(10\mathrm{A}\right)\cos 150°}{{\displaystyle \mathrm{π}×0.50\text{m}}}\right)\hat{\mathrm{i}} \\ \stackrel{→}{\mathrm{B}}=\left(-6.95×{10}^{-6}\mathrm{T}\right)\hat{\mathrm{i}} \end{gathered}$$

To cancel this magnetic field, we have to take the current in the wire $b$into the page along role="math" localid="1663000003728" $-Z$direction.

The current in the wire $b$is given by,

$B=\frac{{\mathrm{μ}}_0{i}_b}{2\mathrm{π}d}$

Where $d$ is the distance from b to a and is given by

role="math" localid="1663000331165" $$\begin{gathered} d=\frac{\sqrt{3}}{2}l \\ =0.43301\text{m} \end{gathered}$$

Hence, the current in$b$is

$$\begin{gathered} i_b=\frac{2\mathrm{π}dB}{{\mathrm{μ}}_0} \\ {i}_b=\frac{{\displaystyle 2\mathrm{π}×0.43301\text{m}×-6.95×{10}^{-6}\text{T}}}{1.26×{10}^{-6}{\displaystyle \raisebox{1ex}{$\mathrm{T}.\mathrm{m}$}\!\left/ \!\raisebox{-1ex}{$\mathrm{A}$}\right.}} \\ {i}_b=15\text{A} \end{gathered}$$

Hence, the current in wire $b$is $15\text{A}.$

To cancel the magnetic field calculated in (a) wire $b$ must carry the current into the page that is along negativez axis.

Hence, the direction of current in wire $b$ is $-Z$ direction.