CET Karnataka Medical
CET - Karnataka Medical Solved Paper-2013
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. \[\therefore \] \[\frac{{{\mu }_{0}}}{4\pi }\frac{2\pi {{I}_{1}}}{R}=\frac{{{\mu }_{0}}}{4\pi }\frac{{{I}_{2}}}{R}\] \[\frac{{{I}_{1}}}{{{I}_{2}}}=\frac{1}{2\pi }\]