In Fig. the circular and the straight parts of the wire are made of same material but have different diameters. The magnetic field at the centre is zero.

The ratio of the diameters of wires of circular and straight parts is

Options

A. `1/(sqrt(2))`
B. `(2sqrt(3))/(pi)`
C. `(3sqrt(3))/(2pi)`
D. `sqrt(2)`

Answer

D

Explanation

From `R=(rhol)/A,(R_(1))/(R_(2))=(l_(1))/(l_(2))(A_(2))/(A_(1))implies(V//I_(1))/(V//I_(2))=(l_(1))/(l_(2))((pi//4)d_(2)^(2))/((pi//4)d_(1)^(2))implies(d_(1))/(d_(2))=sqrt((I_(1))/(I_(2)) (I_(1))/(I_(2)))=sqrt((4pi)/(3sqrt(3)). ( 3 sqrt(3))/(2pi))=sqrt(2)`
Where `I_(1)=2piR(2/3)=4/3pi R` and `I_(2)=2R sin 60^(@)` =`sqrt(3)R`