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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): |
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Answer» The magnetic FIELD at S due to the 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)` |
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