Problem 8 Two long, parallel conductors carry currents in the same direction, as
ID: 2032721 • Letter: P
Question
Problem 8 Two long, parallel conductors carry currents in the same direction, as shown in the figure below. Conductor A carries a current of 150 A and is held firmly in position. Conductor B carries a current Ip and is allowed to slide freely up and down parallel to A) between a set of non-conducting guides. If the mass per unit length of conductor B is 0.190 g/em, what value of current Ig will result in equilibrium when the distance between the two conductors is 2.54 cm? Submit AnswerTries 06 Due Friday March 30 11:59 am (EDT) ??Explanation / Answer
magnetic field due to A at the location of B
BA = uo*IA/(2*pi*r)
magnetic force eerted on B , FB = BA*IB*LB (act up wards)
FB = uo*IA*IB*LB/(2*pi*r)
gravitational force Fg = -m*g ( down wards)
IN equilibrium net force = 0
FB + Fg = 0
uo*IA*IB*LB/(2*pi*r) - mg = 0
uo*IA*IB*LB/(2*pi*r) = m*g
IB = m*g*2*pi*r/(uo*IA*LB)
IB = (m/LB)*g*2*pi*r/(uo*IA)
m/LB = mass per unit length = 0.19 g/cm = 0.19*10^-3/10^-2 kg/m = 0.019 kg/m
IB = 0.019*9.8*2*pi*2.54*10^-2/(4*pi*10^-7*150)
IB = 157.6 A <<<----------ANSWER