(1) Doubling of the current through the wire will result in doubling of its magnetic field.
The magnetic field around a wire is a function of the current I and radial distance r
[tex]B = \frac{\mu\cdot I }{2\pi r}[/tex]
(with mu denoting the magnetic permeability of the medium). So, B is directly proportional to I. The field magnitude will double with the doubled current from 5A to 10A
(2) Using the same formula as in (1), we can see that the magnetic field is inversely proportional to the radial distance from the wire. So, a particle at 20cm will experience half the magnitude compared to a particle at 10cm.
(3) Answer
If a particle with a charge q moves through a magnetic field B with velocity v, it will be acted on by the magnetic force
[tex]F_m = B\cdot v\cdot q[/tex]
So, a particle with charge -2uC will experience a magnetic force of same magnitude but opposite direction (and perpendicular to B) as compared to a particle with a charge of 2uC
If the current in a wire increases from 5 A to 10 A, the magnetic field will double.
If the distance of a charged particle from a wire changes from 10 cm to 20 cm, the magnetic field will be halved.
If the charge of a particle changes from 2 µC to –2µC, the magnitude of the force will be the same but the direction will be different
The given parameters:
The magnetic field around a wire is a function of the current I and radial distance r, given as;
[tex]B = \frac{\mu_o I}{2\pi r} [/tex]
Thus, we can conclude the following;
The force exerted on a particle is calculated as follows;
F = qVB
where;
q is the charge'
If the charge of a particle changes from 2 µC to –2µC, the magnitude of the force will be the same but the direction will be different.
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