question_answer

A wire in the form of a square of side a carries a current i. Then, the magnetic induction at the centre of the square wire is (magnetic permeability of free space \[\frac{x}{a}+\frac{y}{b}=2\]).

A) \[\frac{x}{a}+\frac{y}{b}=\frac{1}{2}\]                                   

B) \[\frac{x}{b}-\frac{y}{a}=2\]

C) \[ax+by=2\]                      

D) \[\frac{{{a}^{2}}}{{{x}^{2}}}+\frac{{{b}^{2}}}{{{y}^{2}}}=1\]

Correct Answer: C

Solution :

Magnetic field due to one side of the square at centre O. \[1.01\times {{10}^{5}}\]\[{{k}_{a}}\]\[\lambda \] Hence, magnetic field at centre due to all sides\[{{k}_{a}},\]