Two wires are hanging from a rectangular ceiling with asupport string on each en
ID: 1743222 • Letter: T
Question
Two wires are hanging from a rectangular ceiling with asupport string on each end of the wire. The current-carrying wires are 2.0m apart. Thecurrent-carrying wires have a mass per unit length of 0.100kg/m. The currents, I, are identical and in the samedirection. a) The support strings are 3.0m long. The support stringshang vertically when there is no current flow. What angle will thesupport strings make from the vertical when a current of 500.0A isflowing in each wire? I found the equation F=ILBsin to try and find ,although I'm not sure if it's even the right equation. I can'tfigure out how to find B or F to plug into the equation because inorder to find the magnetic field I need force, andvisa versa. Help please? Two wires are hanging from a rectangular ceiling with asupport string on each end of the wire. The current-carrying wires are 2.0m apart. Thecurrent-carrying wires have a mass per unit length of 0.100kg/m. The currents, I, are identical and in the samedirection. a) The support strings are 3.0m long. The support stringshang vertically when there is no current flow. What angle will thesupport strings make from the vertical when a current of 500.0A isflowing in each wire? I found the equation F=ILBsin to try and find ,although I'm not sure if it's even the right equation. I can'tfigure out how to find B or F to plug into the equation because inorder to find the magnetic field I need force, andvisa versa. Help please?Explanation / Answer
The distance between the current carrying wires are a =2.0 m apart The current-carrying wires have a mass per unit length of0.100 kg/m. The length of the supporting strings is l = 3.0 m The current flowing in each wire I1=I2= I = 500 A a)The magnetic force on a length l of wire 1 is F1= I1l * B2 Here,l is perpendicular to B2 in this situation,themagnitude of F1 is F1=I1lB2.The magnitude of B2 is givenby the following equation B2= (oI/2a) Here,o= 4 * 10-7NA-2 or F1= I1l * B2=I1l * (oI2/2a) =(oI1I2/2a) * l------------------(1) The direction of F1 is toward wire 2 because l *B2 is in that direction. The gravitational force acting on the wires is F = mg * sin ------------------(2) The gravitational force on the wires is equal to the magneticforce on the wires.Therefore,we get F = F1 or mg * sin =(oI1I2/2a) * l or sin =(oI1I2/2a) * l *(1/mg) Substituting the values in the above equation,we get sin = (4 * 10-7 * 500 * 500/2 *2.0) * 3.0 * (1/0.100 * 3.0 * 9.8) or = 1.46o Therefore,the angle made by the support strings from thevertical when a current of 500.0A is flowing in each wire is = 1.46o. Therefore,the angle made by the support strings from thevertical when a current of 500.0A is flowing in each wire is = 1.46o.Related Questions
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