JEE MAIN - Physics (2020 - 8th January Evening Slot - No. 9)
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 :
_8th_January_Evening_Slot_en_9_1.png)
$${{{\mu _0}I} \over {2\pi R}}\left( {\pi + 1} \right)$$
$${{{\mu _0}I} \over {2\pi R}}\left( {\pi - {1 \over {\sqrt 2 }}} \right)$$
$${{{\mu _0}I} \over {2R}}$$
$${{{\mu _0}I} \over {2\pi R}}\left( {\pi + {1 \over {\sqrt 2 }}} \right)$$
Explanation
_8th_January_Evening_Slot_en_9_2.png)
B = $${{{\mu _0}I} \over {4\pi d}}\left( {\sin 90^\circ - \sin 45^\circ } \right) \otimes $$
+ $${{{\mu _0}I} \over {2R}} \odot $$ + $${{{\mu _0}I} \over {4\pi R}}\left( {\sin 45^\circ + \sin 90^\circ } \right) \odot $$
= $${{{\mu _0}I} \over {2\pi R}}\left( {\sin 45^\circ + \pi } \right)$$
= $${{{\mu _0}I} \over {2\pi R}}\left( {{1 \over {\sqrt 2 }} + \pi } \right)$$
Comments (0)
