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Added picture of simulation of charged particle next to current-carrying wire.
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Mark H
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The field around the wire isn't uniform. When you calculate the force on a charge in a magnetic field, you use the value of the field at the point where the particle is. So, $$F_B = qvB\sin\theta$$ is not just for a constant field. If the field varies from position to position in space, then the force the particle feels will also vary. This equation is introduced with a constant field because the math is easier and because the resulting motion is a simple circle.

So, when you calculate the force of a current-carrying wire on a particle, the resulting force is only for that moment in time. If the particle moves towards or away from the wire, the force on the particle will be different. $F_B = qvB\sin\theta$ is always true, but that doesn't mean the force is constant.

Below is a picture of a proton traveling at $10\ m/s$ next to a wire carrying $1\ A$ of current. The proton starts off at the bottom of the picture traveling upwards parallel to the wire in the same direction as the current. 10 m/s proton next to a 1A wire Notice that as the proton gets closer to the wire, the magnetic field increases, so the turn gets tighter. Once the proton turns $180^o$, it starts moving away from the wire, and the turns become wider.

The field around the wire isn't uniform. When you calculate the force on a charge in a magnetic field, you use the value of the field at the point where the particle is. So, $$F_B = qvB\sin\theta$$ is not just for a constant field. If the field varies from position to position in space, then the force the particle feels will also vary. This equation is introduced with a constant field because the math is easier and because the resulting motion is a simple circle.

So, when you calculate the force of a current-carrying wire on a particle, the resulting force is only for that moment in time. If the particle moves towards or away from the wire, the force on the particle will be different. $F_B = qvB\sin\theta$ is always true, but that doesn't mean the force is constant.

The field around the wire isn't uniform. When you calculate the force on a charge in a magnetic field, you use the value of the field at the point where the particle is. So, $$F_B = qvB\sin\theta$$ is not just for a constant field. If the field varies from position to position in space, then the force the particle feels will also vary. This equation is introduced with a constant field because the math is easier and because the resulting motion is a simple circle.

So, when you calculate the force of a current-carrying wire on a particle, the resulting force is only for that moment in time. If the particle moves towards or away from the wire, the force on the particle will be different. $F_B = qvB\sin\theta$ is always true, but that doesn't mean the force is constant.

Below is a picture of a proton traveling at $10\ m/s$ next to a wire carrying $1\ A$ of current. The proton starts off at the bottom of the picture traveling upwards parallel to the wire in the same direction as the current. 10 m/s proton next to a 1A wire Notice that as the proton gets closer to the wire, the magnetic field increases, so the turn gets tighter. Once the proton turns $180^o$, it starts moving away from the wire, and the turns become wider.

Source Link
Mark H
  • 24.7k
  • 3
  • 63
  • 85

The field around the wire isn't uniform. When you calculate the force on a charge in a magnetic field, you use the value of the field at the point where the particle is. So, $$F_B = qvB\sin\theta$$ is not just for a constant field. If the field varies from position to position in space, then the force the particle feels will also vary. This equation is introduced with a constant field because the math is easier and because the resulting motion is a simple circle.

So, when you calculate the force of a current-carrying wire on a particle, the resulting force is only for that moment in time. If the particle moves towards or away from the wire, the force on the particle will be different. $F_B = qvB\sin\theta$ is always true, but that doesn't mean the force is constant.