A wire carrying current I has the shape as shown in adjoining figure. Linear parts of the wire are very long…

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 :

  1. B→=μ04πIR πi^+2k^
  2. B→=-μ04πIR πi^-2k^
  3. B→=-μ04πIR πi^+2k^
  4. B→=μ04πIR πi^-2k^

Solution


$\begin{aligned} \vec{B}_c & = \vec{B}_1 + \vec{B}_2 + \vec{B}_3 \\ B_1 & = \frac{\mu_0 I}{4 \pi R} (\sin 90 + \sin 0) - k \\ B_1 & = \frac{\mu_0 I}{4 \pi R} (-\hat{k}) = B_3 \end{aligned}$ B due to segment 2
\(B_2=\frac{\mu_0 I}{4 \pi R} \times \pi(-\hat{i})=-\frac{\mu_0 I}{4 R} \hat{i}\)
So B at center \(\vec{B}_c=\vec{B}_1+\vec{B}_2+\vec{B}_3\)
\(B_c=\frac{-\mu_0 I}{4 R}\left(i+\frac{2 \hat{k}}{\pi}\right)=\frac{-\mu_0 I}{4 \pi R}(\pi \hat{i}+2 \hat{k})\)

Asked in: NEET 2015 (Phase 1)

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