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.

Ratio of the currents `I_1 andI_2` flowing through the circular and straight parts is

Options

A. `(sqrt3)/(2pi)`
B. `(2sqrt2)/pi`
C. `(3sqrt3)/(2pi)`
D. `(3sqrt3)/(2sqrt2pi)`

Answer

C

Explanation

(c) 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`