question_answer

In the diagram, are the strength of the currents in the loop and straight conductors respectively. OA=OB=R. The net magnetic field at the canter O is zero. Then the ratio of the currents in the loop and the straight conductors is                 

A)  \[\pi \]      

B)  \[2\pi \]   

C)  \[\frac{1}{\pi }\]     

D)  \[\frac{1}{2\pi }\]

Correct Answer: D

Solution :

  Magnetic field at the centre O due to current \[{{I}_{1}}\]in the loop is \[{{B}_{1}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi {{I}_{1}}}{R}\] Magnetic field at the centre 0 due to current \[{{I}_{2}}\] through straight conductor is \[{{B}_{2}}=\frac{{{\mu }_{0}}}{4\pi }\frac{2{{I}_{2}}}{2R}=\frac{{{\mu }_{0}}}{4\pi }\frac{{{I}_{2}}}{R}\] As the net magnetic field at 0 is zero (given), therefore \[{{B}_{1}}\] and \[{{B}_{2}}\] are equal in magnitude and I opposite in directions. \[\therefore \] \[\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi {{I}_{1}}}{R}=\frac{{{\mu }_{0}}}{4\pi }\frac{{{I}_{2}}}{R}\] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{2\pi }\]