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A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long and parallel to X-axis, while semicircular portion of radius R is lying in Y–Z plane. Magnetic field at point O is
A
$\vec B = \frac{\mu_0 I}{4\pi R}(\pi\hat i + 2\hat k)$
B
$\vec B = -\frac{\mu_0 I}{4\pi R}(\pi\hat i - 2\hat k)$
C
$\vec B = -\frac{\mu_0 I}{4\pi R}(\pi\hat i + 2\hat k)$
D
$\vec B = \frac{\mu_0 I}{4\pi R}(\pi\hat i - 2\hat k)$
Detailed Solution
Field of a straight wire: $B = \frac{\mu_0 I}{4\pi R}(\sin\phi_1 + \sin\phi_2)$
Each semi-infinite straight part gives $\frac{\mu_0 I}{4\pi R}(\sin 90^\circ + \sin 0^\circ) = \frac{\mu_0 I}{4\pi R}$ along $-\hat k$.
$\vec B_L = 2\times\frac{\mu_0 I}{4\pi R}(-\hat k)$
Semicircular part: $\vec B_S = \frac{1}{2}\cdot\frac{\mu_0 I}{2R}(-\hat i) = \frac{\mu_0 I\pi}{4\pi R}(-\hat i)$
$\vec B = \vec B_L + \vec B_S = -\frac{\mu_0 I}{4\pi R}(\pi\hat i + 2\hat k)$
Each semi-infinite straight part gives $\frac{\mu_0 I}{4\pi R}(\sin 90^\circ + \sin 0^\circ) = \frac{\mu_0 I}{4\pi R}$ along $-\hat k$.
$\vec B_L = 2\times\frac{\mu_0 I}{4\pi R}(-\hat k)$
Semicircular part: $\vec B_S = \frac{1}{2}\cdot\frac{\mu_0 I}{2R}(-\hat i) = \frac{\mu_0 I\pi}{4\pi R}(-\hat i)$
$\vec B = \vec B_L + \vec B_S = -\frac{\mu_0 I}{4\pi R}(\pi\hat i + 2\hat k)$