question_answer

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 turm DMND of radius R. AB, BC parts are tangential to circular turn at N and D. Magnetic field at the centre of circle is                                              [JEE MAIN Held on 08-01-2020 Evening]

A) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \pi +1 \right)\]

B)        \[\frac{{{\mu }_{0}}I}{2R}\]

C) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \pi +\frac{1}{\sqrt{2}} \right)\]

D) \[\frac{{{\mu }_{0}}I}{2\pi R}\left( \pi -\frac{1}{\sqrt{2}} \right)\]

Correct Answer: C

Solution :

            \[{{B}_{0}}=\left( {{{\vec{B}}}_{0}} \right)_{1}^{\otimes }+\left( {{{\vec{B}}}_{0}} \right)_{2}^{\odot }+\left( {{{\vec{B}}}_{0}} \right)_{3}^{\odot }+\left( {{{\vec{B}}}_{0}} \right)_{4}^{\odot }\]             \[=-\frac{{{\mu }_{0}}I}{4\pi R}\left( 1-\frac{1}{\sqrt{2}} \right)+\frac{{{\mu }_{0}}I}{2R}+\frac{{{\mu }_{0}}I}{4\pi R}\left( 1+\frac{1}{\sqrt{2}} \right)\] \[\vec{B}_{0}^{\odot }=\frac{{{\mu }_{0}}I}{2\pi R}{{\left( \pi +\frac{1}{\sqrt{2}} \right)}^{\odot }}\]