Three parallel wires are each carrying a 4A current. ; wire A) is 6 mm from wire B) which is 3mm from wire C); The current in wires B and C are out of the paper, while A is into the paper. What is the magnitude and direction of the magnetic field halfway between wires A and B?I have tried using Biot-Savart Law but keep getting it wrong can someone please help? I also have an equation for two parallel wires but how do I relate it to two?
SO YOU HAVE TWO POINTS ON PAPER HAVING DISTANCE 2d. megnetic field zero will be on the mid point of this line. (there will be many points where the megnetic field will be zero but THERE IS ONLY ONE POINT ON PAPER)
You can use the Biot-Savart Law, but it is confusing and kind of a waste of your time and space. For straight wires, someone ALREADY worked out the Biot-Savart Law. See the following link for the result: hyperphysics.phy-astr.gsu.edu/Hba... The formula of interest for us is B = mu0*I/(2*Pi*r) where mu0 is magnetic permeability of free space, I is current, and r is distance from the wire carrying the current. B is the magnetic field due to that PARTICULAR wire. To deal with three wires, use a superposition principle and stack magnetic fields on top of each other. Do be aware of direction: remember the right hand rule. RH rule for magnetic fields in vicinity of wires: point thumb in direction of current, curl fingers to show the magnetic field circulation direction. Use this sign convention: + B is up along page, -B is down along page For Wire A: point of interest is r = d_ab/2 or 0.003 m to the right of wire A Point thumb in to the paper and the finger curl indicates that B_A is downward B_A = -mu0*I_A/(pi*d_ab) For Wire B: point of interest is r = d_ab/2 or 0.003 m to the left of wire B Point thumb out of the paper and the finger curl indicates that B_B is downward B_B = -mu0*I_B/(pi*d_ab) For wire C: point of interest is r = d_ab/2 + d_bc or 0.006 m to the left of wire C Point thumb out of the paper and the finger curl indicates that B_C is downward B_C = -mu0*I_C/(2*pi*(d_ab/2 + d_bc)) Add up: Bnet = B_A + B_B + B_C Bnet = -mu0*I_A/(pi*d_ab) - mu0*I_B/(pi*d_ab) - mu0*I_C/(2*pi*(d_ab/2 + d_bc)) Simplify: Bnet = -mu0/pi*(I_A/d_ab + I_B/d_ab + I_C/(d_ab + 2*d_bc)) data: mu0:=4*Pi*10^(-7) Tesla-m/A; I_A:=4 A; I_B:=4 A; I_C:=4 A; d_ab:=0.006 m;d_bc:=0.003 m; Result: Bnet = -6.667 milliTeslas negative sign indicates downward direction.
