Force between Electric currents

AI Thread Summary
The discussion focuses on calculating the force between electric currents in a setup with fixed and movable wires. The magnetic field strength at the center of the movable wire due to a 9.3 A current in the fixed wire is calculated to be approximately 2 x 10^-4 Tesla. To balance the repulsive force, a mass of about 50.2 mg needs to be added to the movable wire's weight pan. The participants seek guidance on calculating the total magnetic field at a specific point when the currents flow in opposite and same directions. Overall, the thread emphasizes the application of relevant equations to solve the problems presented.
Xgens
Messages
8
Reaction score
0

Homework Statement


[/B]
In a typical experiment to study the force between electric currents, identical fixed and movable wires were separated using a spacer thickness 0.6065cm. The radius of each wire was determined to be 0.162cm and the length of the movable wire was 26.5cm. The apparatus was set up so as to produce a repulsive force when a current of 9.3 A passed through the wires in opposite directions. Assume the magnetic permeability of free space, µ0 = 4p x 10-7 T.m/A.(a) Calculate the magnetic field strength at the center of the movable wire due to the given current through the fixed wire.(b) What mass(in milligrams) would have to be added to the weight pan on the movable wire to balance out the repulsive magnetic force when the given current passes through the fixed wire?(c) Calculate the magnitude of the total magnetic field due to the given current in the wires, along the perpendicular line between the wires at a point y=0.404cm, measured from the surface of the fixed wire when:

(i) the current flows in the opposite direction in the two wires, and

(ii) the current flows in same direction

Homework Equations


B = µ0 * I / 2pi*r
F= ILBsintheta
F/L = µ0 I1 I2 / 2pi * d

The Attempt at a Solution



(a) B = µ0 * 9.3A / 2pi * (0.162cm + 0.162cm + 0.6065cm) -> convert to meter
B = 2 x 10^-4 Tesla

(b) F = ILB sin theta (in this case L x B is perpendicular so it's 90 degree) and F = mg so
mg = ILB
m = ILB / g
m = 9.3A * 0.265m * 2 x 10^-4 T / 9.81
m = 5.02 x 10^-5 kg - > 50.2 mg

(c) I am not quite sure on how to approach this problem, can someone give me a pointer and check to see if i did a & b correctly? Thanks!
 
Physics news on Phys.org
Any ideas or suggestion?
 
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Correct statement about a reservoir with an outlet pipe'
The answer to this question is statements (ii) and (iv) are correct. (i) This is FALSE because the speed of water in the tap is greater than speed at the water surface (ii) I don't even understand this statement. What does the "seal" part have to do with water flowing out? Won't the water still flow out through the tap until the tank is empty whether the reservoir is sealed or not? (iii) In my opinion, this statement would be correct. Increasing the gravitational potential energy of the...
Back
Top