Net magnetic force on a loop parallel to a wire

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A long straight wire carrying a current of 15.8 A is positioned parallel to a square loop with a current of 2.66 A. The net force on the loop is calculated using the equation Force = I2*L*[(μ0*I1)/(2*pi*d)], but the initial calculation yielded an incorrect result. The discussion emphasizes that the magnetic field from the wire decreases with distance, affecting the force on the loop. It is noted that the perpendicular sides of the loop experience equal and opposite forces that cancel each other out, meaning only the parallel segments contribute to the net force. The correct approach involves applying the right-hand rule to determine the direction of the forces and considering the impact of distance on the magnetic field.
Cheezay
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Homework Statement



A long, straight wire carries a current of I1 = 15.8 A. Next to the wire is a square loop with sides L = 1.17 m in length. d = 1.42 m, as shown in the figure below.

The loop carries a current of I2 = 2.66 A in a direction parallel to the wire. Calculate the magnitude of the net force acting on the loop.


Homework Equations



The equation i have used is Force = I2*L*[(μ0*I1)/(2*pi*d)]

The Attempt at a Solution



Because i am looking for force of the loop, here are my numbers plugged in:

Force = 2.66*1.17*[(μ0*15.8A)/(2*3.14*1.42m)]
= 6.93*10^-6N, answer is incorrect


I am stumped.. any help would be greatly appreciated, thanks!
 
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Remember that the straight wire's magnetic field decreases when distance increases. What happens to the force in the parts of the loop that are perpendicular to the wire? Which way are the force vectors pointing?
 
Last edited:
The left perpendicular side of the loop has a force vector pointing up, towards the straight wire, and the right side has a force vector pointing down. Am i using the right equation, or is there more i need to add to it?
 
Cheezay said:
The left perpendicular side of the loop has a force vector pointing up, towards the straight wire, and the right side has a force vector pointing down.

No. Use the right hand rule. Thumb points where current points, index points towards the magnetic field and middle shows the direction of force. You're right, thought, they point in different directions an are equal in magnitude so they cancel each other out. All you have to worry is the parallel parts.

Am i using the right equation, or is there more i need to add to it?

That's the right equation. You just need to remember that distance affects the magnitude of the wire's magnetic field.
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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