question_answer

A cell is connected between the points A and C of a circular conductor ABCD of centre O with angle A\[OC={{60}^{o}}\]. If \[{{B}_{1}}\] and \[{{B}_{2}}\] are the magnitudes of the magnetic fields at O due to the currents in ABC  and ADC  respectively, the ratio \[\frac{{{B}_{1}}}{{{B}_{2}}}\] is [KCET 1999; Pb PET 2000]

A)            0.2

B)            6

C)            1

D)            5

Correct Answer: C

Solution :

                   \[B=\frac{{{\mu }_{0}}}{4\pi }\frac{\theta \,i}{r}\Rightarrow B\propto \theta \,i\]    (but \[\frac{{{i}_{1}}}{{{i}_{2}}}=\frac{{{l}_{2}}}{{{l}_{1}}}=\frac{{{\theta }_{2}}}{{{\theta }_{1}}}\]) \[\Rightarrow \frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\theta }_{1}}}{{{\theta }_{2}}}.\frac{{{i}_{1}}}{{{i}_{2}}}\] So, \[\frac{{{B}_{1}}}{{{B}_{2}}}=\frac{{{\theta }_{1}}}{{{\theta }_{2}}}\times \frac{{{\theta }_{2}}}{{{\theta }_{1}}}\] \[\Rightarrow {{B}_{1}}={{B}_{2}}\]