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/(sqrt2)`
B. `(2sqrt3)/pi`
C. `(3sqrt3)/(2pi)`
D. `sqrt2`

Answer

D

Explanation

(d) If `I_1 and I_2` be the currents in circular and straight parts,
respectively, and `B_1, B_2` the magnetic field due to them, then
`vecB_1=(mu_0I_1)/(2R)xx2/3 =(mu_0I_1)/(3R)`
`B_2=(mu_0I_2)/(4pi[Rcos60^@])[2sin60^@]=(sqrt3mu_0I_2)/(2piR)`
For the total field at 'O' to be zero,
`(mu_0I_1)/(3R)=(sqrt3mu_0I_2)/(2piR) implies (I_1)/(I_2)=(3sqrt3)/(2pi)`

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((l_1I_1)/(l_2I_2))`
where `l_1=2piR(2/3)=4/3piR and l_2=2Rsin60^@=sqrt3R`
`implies (d_1)/(d_2)=sqrt((4piR)/(3sqrt3R) (3sqrt3)/(2pi))=sqrt2`