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 :

 

  • Option 1)

    B= \frac{\mu_0 I}{4 \pi R}(\pi \hat{i}+ 2\hat{k})

  • Option 2)

    B= -\frac{\mu_0 I}{4 \pi R}(\pi \hat{i}- 2\hat{k})

  • Option 3)

    B= -\frac{\mu_0 I}{4 \pi R}(\pi \hat{i}+ 2\hat{k})

  • Option 4)

    B= \frac{\mu_0 I}{4 \pi R}(\pi \hat{i}- 2\hat{k})

 

Answers (1)

As discussed in

For Infinite Length -

phi_{1}=phi_{2}=90^{circ}

B=frac{mu_{o}}{4pi}:frac{2i}{r}:

- wherein

 

 Parallel wires 1 and 3 are semi infinite, so magnetic field at 0 due to them

\vec{B_{1}}=\vec{B_{2}}=\frac{-\mu _{0I}}{4\pi R}\hat{k}

Magnetic field at 0 due to semicircular arc in yz plane is given by

\vec{B_{2}} = \frac{-\mu _{0}I}{4R}\hat{i}

\vec{B}=\vec{B_{1}}+\vec{B_{2}}+\vec{B_{3}}=\frac{-\mu _{0}I}{4\pi R}\hat{K}-\frac{\mu _{0}I}{4R}\hat{i}-\frac{\mu _{0}I}{4\pi R}\hat{K}

\vec{B}=-\frac{\mu _{I}}{4\pi R} (\pi \hat{i}+2\hat{k})

 

 


Option 1)

B= \frac{\mu_0 I}{4 \pi R}(\pi \hat{i}+ 2\hat{k})

Option 2)

B= -\frac{\mu_0 I}{4 \pi R}(\pi \hat{i}- 2\hat{k})

Option 3)

B= -\frac{\mu_0 I}{4 \pi R}(\pi \hat{i}+ 2\hat{k})

Option 4)

B= \frac{\mu_0 I}{4 \pi R}(\pi \hat{i}- 2\hat{k})

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