The drawing shows two long, straight wires that are suspended from the ceiling. The mass per unit of length of each wire is 0.050kg/m. Each of the four strings suspending the wires has a length of 1.2m. When the wires carry identical currents in opposite directions, the angel between the strings holding the two wires is 15 degrees. What is the current in each wire?? Im not sure what equations to use for this problem.
Look at it this way a current I2 causes a magnetic field. if a current I1 is in a magnetic field then a force acts on it. F=I1*B*dl where B=I2*(mu0)/(2*pi*d) So the force between those two wires is given by this equation F=I1*I2*(mu0)*dl/(2*pi*d) mu0=4*pi*10power7 d the distance between them In this case I1=I2 dl which shows the lenght of the wire in the magnetic field equals 1 because you do not care about it's lenght from here on it is rather simple I hope
Wiring that previous could desire to get replaced. Insulators wreck down over the years. In Edison's day insulation became into, rope and tar, or fabric, with glass insulators to guard lumber. you like a rectifier to create pulse ac (see image voltaic Panel grant) i hit upon it not straight forward to have faith anybody might grant DC power. The generator could desire to be on the threshold of factor to apply, do to severe voltage drop in DC cutting-edge over long distance. through fact of this AC is used now. till that's an Edison Museum image voltaic power is DC too. I advise studying image voltaic power for ideal DC wiring practices. I actual have been a electrician my total existence and know little or no approximately DC wiring and contraptions. in case you presently have electrical energy, get multimeter attempt for ac voltage first. then DC to steer away from unfavorable the meter, to confirm what power you have
You need two basic equation: 1° Force acting on conductor in magnetic field: F = B I λ ..........(1) where B is magnetic field (also called magnetic induction) I is current λ is length of wire 2° Magnetic field near straight wire carrying current I B = μo I / (2πd) ..........(2) where d is distance from the wire μo is the magnetic permeability constant of vacuum (approx. same for air) μo= 4π*10^-7 Tm/A When there are two parallel wires on distance d carrying currents I1 and I2, each of them produce magnetic field causing forces acting on wires. Magnetic field from second wire creates force on first wire: F1 = B2 * I1 * λ Magnetic field from first wire creates force on second wire: F2 = B1 * I2 * λ These forces are equal in magnitude: F = μo I1 I2 λ / (2πd) ..........(3) When currents flow in same direction forces are attractive, and for opposite direction forces are repulsive. When I1=I2, F = μo I? λ / (2πd) ..........(4) Each wire, suspended on strings, decline from vertical position for angle θ=15°/2 = 7.5°, so that the angle between the strings holding the two wires is 2θ=15°. We find F from condition of static equilibrium of torques (relative to point where strings are attached to ceiling): F L cos θ = G L sin θ ..........(5) where G is weight of wire; G = m'gλ where m' is the mass per unit of length of each wire. L is length of strings from (5) we get relation F = G tan θ ..........(6) or F = m' g λ tan θ ..........(7) and from (4) and (7) : μo I? λ / (2πd) = m' g λ tan θ ..........(8) λ on both sides cancels, and we find current I as I = √(2 π d m' g tan θ / μo) ..........(9) You didn't say if there was some initial distance between wires, so we'll assume they are suspended in same points. In that case d = 2L sinθ ..........(10) and finally I = 2 √(π L m' g sinθ tan θ / μo) ..........(11) I = √[1.2 * 0.05 * 9.81 * sin 7.5° * tan 7.5° / 10^-7 ] I = 318.034 A
