If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f

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If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f

rom a wire changes from 10 cm to 20 cm, what happens to its magnetic field? If the charge of a particle changes from 2 µC to –2µC, what happens to the force exerted on that particle?

Physics

2 answers:

dsp73 · Answer 1 · 2022-05-02T11:44:17+03:00

1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

where $\mu_0$ is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

2. The magnitude of the magnetic field halves

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

$I=\frac{\mu_0 I}{2 \pi r}$

We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

3. The force reverses direction

Explanation: the force exerted on a charged particle in a magnetic field is:

$F=qvB sin \theta$

where q is the charge, v is the speed of the particle, B is the magnetic field intensity and $\theta$ the angle between the direction of v and B. If the charge of the particle is switched from 2 µC to –2µC, the magnitude of the force does not change (because the absolute value of q does not change), however the charge q gets a negative sign (-), so the sign of the force changes and gets a negative sign too, so the force reverses direction.