question_answer

Three circular concentric wires of radii a, 2a and 3a are carrying current \[3I,\,\,2I,\]and\[I\]in same manner. The magnetic field at the common centre is

A)  \[\frac{13{{\mu }_{0}}I}{6a}\]                   

B)  \[\frac{{{\mu }_{0}}I}{6a}\]

C)  \[\frac{{{\mu }_{0}}I}{a}\]                                          

D)  None of these

Correct Answer: A

Solution :

                \[B={{B}_{1}}+{{B}_{2}}+{{B}_{3}}\] \[=\frac{{{\mu }_{0}}(3I)}{2a}+\frac{{{\mu }_{0}}(2I)}{(2a)}+\frac{{{\mu }_{0}}I}{2(3a)}\] \[=\frac{3{{\mu }_{0}}I}{2a}+\frac{{{\mu }_{0}}I}{2a}+\frac{{{\mu }_{0}}I}{6a}\] \[=\frac{9{{\mu }_{0}}I+3{{\mu }_{0}}I+{{\mu }_{0}}I}{6a}=\frac{13{{\mu }_{0}}I}{6a}\]