Question 187: 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 :

Accepted Answer

B = $${{{\mu _0}I} \over {4\pi d}}\left( {\sin 90^\circ  - \sin 45^\circ } \right) \otimes $$
+ $${{{\mu _0}I} \over {2R}} \odot $$ + $${{{\mu _0}I} \over {4\pi R}}\left( {\sin 45^\circ  + \sin 90^\circ } \right) \odot $$
= $${{{\mu _0}I} \over {2\pi R}}\left( {\sin 45^\circ  + \pi } \right)$$
= $${{{\mu _0}I} \over {2\pi R}}\left( {{1 \over {\sqrt 2 }} + \pi } \right)$$

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: