Force on wire due to magnetic field

In summary, the conversation is about finding the force on a current-carrying wire in a uniform magnetic field. The equation used is F = ILBsinθ and the attempt at a solution involves finding the angle between the wire and the magnetic field and plugging in the given values. The final answer is incorrect due to a wrong sign, but the issue has been resolved.
  • #1
nickclarson
32
0

Homework Statement



A straight segment of a current-carrying wire has a current element IL where I = 2.70 A and L = 2.60 cm i + 4.40 cm j. The segment is in a region with a uniform magnetic field given by 1.36 T i. Find the force on the segment of wire. (Give the x, y, and z components.)

Homework Equations



[tex]\theta = tan^{-1}(y/x)[/tex]

[tex]F = ILBsin\theta[/tex]

The Attempt at a Solution



Well I am pretty sure there will be no x or y component because the force will be perpendicular in the z direction.

To find the z component I found the angle between the wire and the mag field using:

[tex]tan^{-1}(4.4/2.6) = 59.421[/tex]

After that I thought you are just supposed to plug everything in:

[tex]2.7A * .05111m * 1.36 * sin(59.421) = .1615N[/tex]

But it's wrong! I'm not sure what I am doing wrong, the book I am using isn't very good ad explaining. I'm sure it's something simple that I am just overlooking.

Thanks,
Nick
 
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  • #2
CANCEL that. I just had the wrong sign. I understand why it should be negative now. This thread can be deleted.
 
  • #3




Hello Nick,

Your approach seems correct, but there are a few small errors in your calculations. Firstly, when finding the angle using tan^{-1}, make sure to convert the lengths to meters (2.60 cm = 0.0260 m and 4.40 cm = 0.0440 m). This will give you a slightly different angle, around 59.65 degrees.

Secondly, when calculating the z-component of the force, make sure to use the correct length for L. In this case, L is given as 2.60 cm i + 4.40 cm j, so the z-component will be multiplied by 4.40 cm, not 0.05111 m. This will give you a z-component of 0.281 N.

Overall, your approach was correct, but just make sure to double check your calculations and units. I hope this helps!
 

FAQ: Force on wire due to magnetic field

What is the force on a wire due to a magnetic field?

The force on a wire due to a magnetic field is known as the Lorentz force, and it is given by the equation F = qv x B, where q is the charge of the particle, v is its velocity, and B is the magnetic field. This force is perpendicular to both the velocity of the particle and the direction of the magnetic field.

How does the direction of the magnetic field affect the force on a wire?

The direction of the magnetic field determines the direction of the force on a wire. If the magnetic field is parallel to the wire, the force will be zero. If the magnetic field is perpendicular to the wire, the force will be at its maximum.

What is the relationship between the strength of the magnetic field and the force on a wire?

The force on a wire is directly proportional to the strength of the magnetic field. This means that as the strength of the magnetic field increases, the force on the wire will also increase.

Can the force on a wire be controlled?

Yes, the force on a wire can be controlled by changing the strength or direction of the magnetic field. This can be done by adjusting the current in an electromagnet or by changing the orientation of a permanent magnet.

What is the practical application of understanding the force on a wire due to a magnetic field?

Understanding the force on a wire due to a magnetic field is crucial in the design and operation of electric motors, generators, and other electromagnetic devices. It is also used in various scientific experiments, such as determining the charge-to-mass ratio of a particle or studying the behavior of charged particles in a magnetic field.

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