An equilateral triangle ABC of side length a is made. The sides AB and AC have resistance per unit length `beta`, and side BC has resistasnce per unit length `2beta`. An ideal battery of EMF V is now connected across BC as shown. The magnetic field intensity at the centrroid of the triangle ABC is `n((mu_(0)V)/(4pi beta a^(2)))`, then the value of n is _______. (Consider only the field due to sides AB, BC and AC of the triangle, and neglect the field due to other wires)

Answer

3

Explanation

`R_(BAC)=beta(2a)=2betaa,R_(BC)=2beta(a)=2beta aimpliesI_(BAC)=I-V/(2beta),I_(Bc)=I=V/(2beta)`
Field at centroid due to AB , `B_(0)=(mu_(0)I)/(4pi(a/(2sqrt(3))))(sin +60^(@)+sin 60^(@))`
`B_(0)=(3mu_(0)I)/(4pia)` (into the plane), Net field at centroid `=2B_(0)-B_(0),=B_(0)=(3mu_(0)V)/(4pibeta a^(2))`