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Two infinitely long wires are placed at $(1 \mathrm{~cm}$, $1 \mathrm{~cm})$ and $(+1 \mathrm{~cm},-1 \mathrm{~cm})$ with $1 \mathrm{~A}$ current in each and in the same directions perpendicular to $x y$-plane. let the magnetic field due to these current carrying wires at the origin be $\mathbf{B}$. If $B_0$ is the magnitude of the field if only one of them was present. then $\frac{|\mathbf{B}|}{B_0}$ is
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Verified Answer
The correct answer is:
$\sqrt{2}$
The given situation is shown below
Wires are located at position $A$ and $B$
$$
\begin{aligned}
O A & =O B=\sqrt{1^2+1^2}=\sqrt{2} \mathrm{~cm} \\
& =\sqrt{2} \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Magnitude of net magnetic field due to both wire having current $1 \mathrm{~A}$ is given as
$$
\begin{aligned}
|\mathbf{B}| & =\sqrt{\left[\frac{\mu_0 I}{2 \pi(O A)}\right]^2+\left[\frac{\mu_0 I}{2 \pi(O B)}\right]^2} \\
& =\sqrt{\left(\frac{\mu_0 \times 1}{2 \pi \sqrt{2}}\right)^2+\left(\frac{\mu_0 \times 1}{2 \pi \sqrt{2}}\right)^2}=\frac{\mu_0}{2 \sqrt{2} \pi} \sqrt{2}=\frac{\mu_0}{2 \pi} \\
\Rightarrow & |\mathbf{B}|=\frac{\mu_0}{2 \pi}
\end{aligned}
$$
Magnitude of magnetic field due to individual wire $O A$,
$$
\begin{aligned}
& B_0=\frac{\mu_0 I}{2 \pi(O A)}=\frac{\mu_0 \times I}{2 \pi \sqrt{2}}=\frac{\mu_0}{2 \sqrt{2} \pi} \\
\Rightarrow \quad & \frac{|\mathbf{B}|}{B_0}=\frac{\mu_0 / 2 \pi}{\mu_0 / 2 \sqrt{2} \pi}=\sqrt{2}
\end{aligned}
$$
Wires are located at position $A$ and $B$
$$
\begin{aligned}
O A & =O B=\sqrt{1^2+1^2}=\sqrt{2} \mathrm{~cm} \\
& =\sqrt{2} \times 10^{-2} \mathrm{~m}
\end{aligned}
$$
Magnitude of net magnetic field due to both wire having current $1 \mathrm{~A}$ is given as
$$
\begin{aligned}
|\mathbf{B}| & =\sqrt{\left[\frac{\mu_0 I}{2 \pi(O A)}\right]^2+\left[\frac{\mu_0 I}{2 \pi(O B)}\right]^2} \\
& =\sqrt{\left(\frac{\mu_0 \times 1}{2 \pi \sqrt{2}}\right)^2+\left(\frac{\mu_0 \times 1}{2 \pi \sqrt{2}}\right)^2}=\frac{\mu_0}{2 \sqrt{2} \pi} \sqrt{2}=\frac{\mu_0}{2 \pi} \\
\Rightarrow & |\mathbf{B}|=\frac{\mu_0}{2 \pi}
\end{aligned}
$$
Magnitude of magnetic field due to individual wire $O A$,
$$
\begin{aligned}
& B_0=\frac{\mu_0 I}{2 \pi(O A)}=\frac{\mu_0 \times I}{2 \pi \sqrt{2}}=\frac{\mu_0}{2 \sqrt{2} \pi} \\
\Rightarrow \quad & \frac{|\mathbf{B}|}{B_0}=\frac{\mu_0 / 2 \pi}{\mu_0 / 2 \sqrt{2} \pi}=\sqrt{2}
\end{aligned}
$$
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