A triangular current carrying loop is placed in the X-Y plane with its vertices at `P(0,(asqrt(3))/2),Q(-a/2,0)` and `R(a/2,0)`.
The loop carries a current l in the direction `PtoQtoRtoP`. Let the point S have the coordinates `(0,(asqrt(3))/2)`
Choose the correct option(s):

Options

A. The magnetic field at S due to the side PQ is in the –Z direction
B. The magnetic field at S due to the side PQ and the magnetic field at S due to the side RP has the same magnitude
C. The net magnetic field at S is in the –Z direction
D. The net magnetic field at S has magnitude `(mu_(0)I)/(2pia)(2/(sqrt(3))-1)`

Answer

B, C, D

Explanation

Field at S due to side PQ is in the +Z direction.
The perpendicular distance of S from PQ,
`r_(PQ)=(asqrt(3))/2`
So field due ot PQ `B_(PQ)=(mu_(0)I)/(4pir_(PQ))(sin 120^(@)+sin (-30^(@)))=(mu_(0)I)/(4pia)(1-1/(sqrt(3)))`
Field at S due to side RP is also in the +Z direction and has the same magnitude due to symmetry.
Now, field at S due to side QR is in the –Z direction.
The perpendicular distance of S from QR,
`r_(QR)=(asqrt(3))/2`
So field due to QR` B_(QR)=(mu_(0)I)/(4pir_(QR))(sin 30^(@)+sin 30^(@))=(mu_(0)I)/(2sqrt(3)pia)`
Net field at S `vecB_("net")=vecB_(PQ)+vecB_(QR)+veB_(RP)=[(mu_(0)I)/(2pia)(2/(sqrt(3))-1)](-hatk)`