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A long straight wire carrying current $16 \mathrm{~A}$ is bent at $90^{\circ}$ such that half of the wire lies along the positive $x$-axis and other half lies along the positive $y$-axis. What is the magnitude of the magnetic field at the point, $\mathbf{r}=(-2 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}) \mathrm{mm}$ ? (Assume, $\frac{\mu_0}{4 \pi}=10^{-7} \mathrm{H} \mathrm{m}^{-1}$ )
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Verified Answer
The correct answer is:
$0.8 \mathrm{mT}$
According to the question, a long straight wire is bent at $90^{\circ}$ as shown in the figure,
Where, current flowing through the wire, $I=16 \mathrm{~A}$
$$
\mathbf{r}=\mathrm{OP}=(-2 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}) \mathrm{mm}
$$
Magnitude of the magnetic field at point $P$ due to current carrying wire along $x$-axis is zero because pont $P$ lies in the direction of conducting wire along $x$-axis.
$\therefore$ Magnitude of the magnetic field due to current carrying wire along $y$-axis at point $P$ is given by
$$
B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}
$$
Putting the given values in above relation, we get
$$
\begin{aligned}
& =\frac{\mu_0}{4 \pi} \times \frac{16}{2 \times 10^{-3}} \quad\left[\because r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\right] \\
& =10^{-7} \times 8 \times 10^3=0.8 \times 10^{-3} \mathrm{~T}=0.8 \mathrm{mT}
\end{aligned}
$$
Where, current flowing through the wire, $I=16 \mathrm{~A}$
$$
\mathbf{r}=\mathrm{OP}=(-2 \hat{\mathbf{i}}+0 \hat{\mathbf{j}}) \mathrm{mm}
$$
Magnitude of the magnetic field at point $P$ due to current carrying wire along $x$-axis is zero because pont $P$ lies in the direction of conducting wire along $x$-axis.
$\therefore$ Magnitude of the magnetic field due to current carrying wire along $y$-axis at point $P$ is given by
$$
B=\frac{\mu_0}{4 \pi} \cdot \frac{I}{r}
$$
Putting the given values in above relation, we get
$$
\begin{aligned}
& =\frac{\mu_0}{4 \pi} \times \frac{16}{2 \times 10^{-3}} \quad\left[\because r=2 \mathrm{~mm}=2 \times 10^{-3} \mathrm{~m}\right] \\
& =10^{-7} \times 8 \times 10^3=0.8 \times 10^{-3} \mathrm{~T}=0.8 \mathrm{mT}
\end{aligned}
$$
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