Force between Electric currents

In summary, the conversation discusses a typical experiment to study the force between electric currents using fixed and movable wires separated by a spacer. The magnetic field strength at the center of the movable wire due to a given current through the fixed wire is calculated using the equations B = µ0 * I / 2pi*r and F= ILBsintheta. The mass needed to balance out the repulsive magnetic force is also calculated using the equation F = ILB and F=mg. The magnitude of the total magnetic field at a point between the wires is to be calculated when the currents flow in opposite and same directions. No solution is provided for this part.
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Xgens
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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!
 
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Any ideas or suggestion?
 

Related to Force between Electric currents

1. What is the force between two parallel electric currents?

The force between two parallel electric currents is attractive when the currents are in the same direction and repulsive when they are in opposite directions. The magnitude of the force is directly proportional to the product of the two currents and inversely proportional to the distance between them.

2. How is the direction of the force between electric currents determined?

The direction of the force between electric currents is determined by the right hand rule. If the right hand is curled in the direction of the first current, with the fingers pointing in the direction of the second current, the thumb will point in the direction of the force.

3. How does the presence of a magnetic field affect the force between electric currents?

The presence of a magnetic field can either strengthen or weaken the force between electric currents, depending on the direction of the magnetic field. If the magnetic field is parallel to the currents, it will strengthen the force, and if it is perpendicular, it will weaken the force.

4. What is the relationship between distance and the force between electric currents?

The force between electric currents is inversely proportional to the square of the distance between them. This means that as the distance increases, the force decreases exponentially.

5. How is the force between electric currents affected by the medium they are in?

The force between electric currents is affected by the medium they are in through the permeability of the medium. The force is stronger in materials with higher permeability, such as iron, and weaker in materials with lower permeability, such as air.

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