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Three long, straight, parallel wires carrying different currents are arranged as shown in the diagram. In the given arrangement, let the net force per unit length on the wire ' $C$ ' be $\vec{F}$. If the wire ' $B$ ' is removed without disturbing the other two wires, then the force per unit length on wire ' $\mathrm{A}$ ' is
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The correct answer is:
$-3 \overrightarrow{\mathrm{F}}$
Force on one wire. due to the other current carrying wive is $\quad F=\frac{\mu_0^2 i_1 i_2 \times e}{4 \pi x}$
Given $\left(\frac{F}{l}\right)_c=\vec{F}$
$\begin{aligned} & =+\frac{\mu_0}{4 d} 2(2 i)(i)+\left(-\frac{\mu_0}{4 \pi(2 d)} 2(2 i)(3 i)\right) \\ & =\frac{\mu_0 i^2}{4 \pi d}(-6+4) \Rightarrow \vec{F}=-\frac{2 \mu_0 i^2}{4 \pi d}\end{aligned}$
Now we need to calculate force per unit length on wire $A$ if $B$ is removed
$\begin{aligned} \Rightarrow\left(\frac{F}{l}\right)_A & =-\frac{\mu_0}{4 \pi} \frac{2(2 i)(3 i)}{2 d} \\ \left(\frac{F}{l}\right)_A & =3\left(\frac{2 \mu_0 i^2}{4 \pi}\right) \\ \left(\frac{F}{e}\right)_A & =-3 \vec{F} \end{aligned}$
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