Question 196: Magnetism and Matter

Question

A very long wire ABDMNDC is shown in
figure carrying current I. AB and BC parts are
straight, long and at right angle. At D wire
forms a circular turn DMND of radius R. AB,
BC parts are tangential to circular turn at N and
D. Magnetic field at the centre of circle is :
![](https://imagex.cdn.examgoal.net/7k9k2k5hh81cs/6a89aaf4-ff8c-4f2c-9e9c-afa2fe719317/96f7a1c0-38c7-11ea-92dd-4b9c3519996f/file-7k9k2k5hh81ct-249w.jpg)

Accepted Answer

B = ${{{\mu _0}I} \over {4\pi d}}\left( {\sin 90^\circ  - \sin 45^\circ } \right) \otimes $
+ ${{{\mu _0}I} \over {2R}} \odot $ + ${{{\mu _0}I} \over {4\pi R}}\left( {\sin 45^\circ  + \sin 90^\circ } \right) \odot $
= ${{{\mu _0}I} \over {2\pi R}}\left( {\sin 45^\circ  + \pi } \right)$
= ${{{\mu _0}I} \over {2\pi R}}\left( {{1 \over {\sqrt 2 }} + \pi } \right)$
![](https://imagex.cdn.examgoal.net/7kd4dk6a3x01i/43ff05da-ae5b-4b04-9004-9b2ee7a3bb50/4834d160-4886-11ea-b62d-51ccc5d4afb6/file-7kd4dk6a3x01j-267w.jpg)

Answer

${{{\mu _0}I} \over {2\pi R}}\left( {\pi  + 1} \right)$

Answer

${{{\mu _0}I} \over {2\pi R}}\left( {\pi  - {1 \over {\sqrt 2 }}} \right)$

Answer

${{{\mu _0}I} \over {2R}}$

Answer

${{{\mu _0}I} \over {2\pi R}}\left( {\pi  + {1 \over {\sqrt 2 }}} \right)$

Magnetism and Matter

A very long wire ABDMNDC is shown in figure carrying current I. AB and BC parts are straight, long and at right angle. At D wire forms a circular turn DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is :

Choose the correct answer: