A cell is connected between the points A and C of a circular conductor ABCD of centre `'O'.angle AOC = 60^(@)`. 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 `(B_(1))/(B_(2))`, is

Options

A. `0.2`
B. 6
C. 1
D. 5

Answer

C

Explanation

`B_(1) = (mu_(0)I_(1))/(4pi R) . (theta_(1)) = (mu_(0) I_(1))/(4pi R) ((5pi)/(3))`
`B_(2) = (mu_(0) I_(2))/(4pi R) (theta_(2)) = (mu_(0) I_(2))/(4pi R) ((pi)/(3)) . (B_(1))/(B_(2)) = ((I_(1))/(I_(2))) xx 5`
Now , `(I_(1))/(I_(2)) = (I_(2))/(I_(1)) = (R xx pi //3)/(R xx 5pi //3) = (1)/(5)` [ as ` I prop (1)/("Resistance") prop (1)/(l)`]
`(B_(1))/(B_(2)) =1 implies B_(1) : B_(2) = 1 : 1`