A very long wire A B D M N D C is shown in figure carrying current I.
Question:

A very long wire $A B D M N D C$ is shown in figure carrying current $I$. $A B$ and $B C$ parts are straight, long and at right angle. At $D$ wire forms a circular turn $D M N D$ of radius $R$.

$A B, B C$ parts are tangential to circular turn at $N$ and $D$. Magnetic field at the centre of circle is:

  1. (1) $\frac{\mu_{0} I}{2 \pi R}\left(\pi+\frac{1}{\sqrt{2}}\right)$

  2. (2) $\frac{\mu_{0} I}{2 \pi R}\left(\pi-\frac{1}{\sqrt{2}}\right)$

  3. (3) $\frac{\mu_{0} I}{2 \pi R}(\pi+1)$

  4. (4) $\frac{\mu_{0} I}{2 R}$


Correct Option: 1

Solution:

$B_{0}=B_{1}+B_{2}+B_{3}+B_{4}$

$=\frac{\mu_{0} I}{4 \pi R}\left[\sin 90^{\circ}-\sin 45^{\circ}\right]+\frac{\mu_{0} I}{2 R}+\frac{\mu_{0} I}{4 \pi R}$

$\left[\sin 45^{\circ}+\sin 90^{\circ}\right]$

$=-\frac{\mu_{0} I}{4 \pi R}\left(1-\frac{1}{\sqrt{2}}\right)+\frac{\mu_{0} I}{2 R}+\frac{\mu_{0} I}{4 \pi R}\left(1+\frac{1}{\sqrt{2}}\right)$

$\overrightarrow{\mathrm{B}_{0}^{\odot}}=\frac{\mu_{0} \mathrm{I}}{2 \pi \mathrm{R}}\left(\pi+\frac{1}{\sqrt{2}}\right)^{\odot}$

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