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