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(C) The given situation is shown in the figure. The wire consists of three parts: two semi-infinite straight wires parallel to the $X$-axis and a semi

Vedclass · Accepted Answer

$\overrightarrow {B} = - \frac{{\mu _0}I}{{4\pi R}}\left( {\pi \hat i + 2\hat k} \right)$

Vedclass · Answer

(C) The given situation is shown in the figure. The wire consists of three parts: two semi-infinite straight wires parallel to the $X$-axis and a semicircular arc in the $Y-Z$ plane.$1$. For the two semi-infinite straight wires (labeled $1$ and $3$): The magnetic field at $O$ due to a semi-infinite wire is $B = \frac{\mu_0 I}{4 \pi R}$. Using the right-hand rule,both wires produce a magnetic field in the $-\hat{k}$ direction at point $O$. Thus,$\vec{B}_{1} = \vec{B}_{3} = -\frac{\mu_{0} I}{4 \pi R} \hat{k}$.$2$. For the semicircular arc (labeled $2$): The magnetic field at the center of a semicircular arc of radius $R$ is $B = \frac{\mu_0 I}{4 R}$. Using the right-hand rule,the field is directed in the $-\hat{i}$ direction. Thus,$\vec{B}_{2} = -\frac{\mu_{0} I}{4 R} \hat{i}$.$3$. The net magnetic field at point $O$ is the vector sum: $\vec{B} = \vec{B}_{1} + \vec{B}_{2} + \vec{B}_{3}$.Substituting the values: $\vec{B} = -\frac{\mu_{0} I}{4 \pi R} \hat{k} - \frac{\mu_{0} I}{4 R} \hat{i} - \frac{\mu_{0} I}{4 \pi R} \hat{k}$.Factoring out $-\frac{\mu_{0} I}{4 \pi R}$: $\vec{B} = -\frac{\mu_{0} I}{4 \pi R} (\pi \hat{i} + 2 \hat{k})$.

$A$ wire carrying current $I$ has the shape as shown in the adjoining figure. The linear parts of the wire are very long and parallel to the $X$-axis,while the semicircular portion of radius $R$ lies in the $Y-Z$ plane. The magnetic field at point $O$ is:

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