question_answer

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

A)  \[\frac{13{{\mu }_{0}}l}{6a}\]

B)  \[\frac{{{\mu }_{0}}l}{6a}\]

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

D)  none of these

Correct Answer: A

Solution :

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