A cell is connected between the points A and Cofa circular conductor ABCDA of centre O. `angleAOC=60^(@)`. If `B_(1)`, and `B_(2)`, are the magnitudes of the magnetic fields at O due to currents in ABC and ADC respectively, the ratio `B_(1)"/"B_(2)`, is

Options

A. 6
B. 5
C. 1
D. `1"/"5`

Answer

C

Explanation

Let `l_(1),l_(2)` be the lengths of the two parts ABC and ADC of the lengths of the two parts ABC and ADC of the conductor and `rho` be resistance per unit length of the conductor .
The resistance of the part ABC will be
`R_(1)=rhol_(2)"/"A`
The resistance of the part ABC will be
`R_(2)=rhol_(2)"/"A`
Potential difference across AC must be equal `I_(1)R_(1)=I_(2)rhol_(2)`
or `I_(1)rhol_(2)"/"A=I_(2)rhol_(2)"/"A` " "0r " " `I_(1)l_(1)=I_(2)l_(2)`
Magnetic field at O due to a cuttent `I_(1)` flowing in ABC is given by
`B_(1)=(mu_(0)I_(1)theta_(1))/(4pir)=(mu_(0)I_(1)l_(1))/(4pir^(2))" "(Asl_(1)=theta_(1)r)`
Magnetic field at O due to current `I_(2)` flowing in ADC is given by
`B_(2)=(mu_(0)I_(2)theta_(2))/(4pir)=(mu_(0)I_(2)l_(2))/(4pir^(2))" "(Asl_(2)=theta_(2)r)`
`:. (B_(1))/(B_(2))=1`