A) \[\frac{\sqrt{5}{{\mu }_{0}}I}{2\,R}\]
B) \[\frac{3{{\mu }_{0}}I}{2\,R}\]
C) \[\frac{{{\mu }_{0}}I}{2\,R}\]
D) \[\frac{{{\mu }_{0}}I}{R}\]
Correct Answer: A
Solution :
Magnetic field induction due to vertical loop at the centre O is \[{{B}_{1}}=\frac{{{\mu }_{0}}l}{2R}\] It acts in horizontal direction. Magnetic field induction due to horizontal loop at the centre O is. \[{{B}_{2}}=\frac{{{\mu }_{0}}2l}{2R}\] It acts in vertically upward direction. As 6, and 63 are perpendicular to each other, therefore the resultant magnetic field induction at the centre O is \[{{B}_{net}}=\sqrt{B_{1}^{2}+B_{2}^{2}}=\sqrt{{{\left( \frac{{{\mu }_{0}}l}{2R} \right)}^{2}}+{{\left( \frac{{{\mu }_{0}}2l}{2R} \right)}^{2}}}\] \[=\,\,\,\,\frac{{{\mu }_{0}}l}{2R}\sqrt{{{l}^{2}}+{{2}^{2}}}=\frac{\sqrt{5}{{\mu }_{0}}l}{2R}\]You need to login to perform this action.
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