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
2 answers:
1. The magnitude of the magnetic field doubles
Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:
where 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:
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:
where q is the charge, v is the speed of the particle, B is the magnetic field intensity and 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.
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Answer:
t_{out} = t_{in}, t_{out} =
Explanation:
This in a relative velocity exercise in one dimension,
let's start with the swimmer going downstream
its speed is
The subscripts are s for the swimmer, r for the river and g for the Earth
with the velocity constant we can use the relations of uniform motion
= D /
D = v_{sg1} t_{out}
now let's analyze when the swimmer turns around and returns to the starting point
= D /
D = v_{sg 2} t_{in}
with the distance is the same we can equalize
t_{out} = t_{in}
t_{out} = t_{in}
This must be the answer since the return time is known. If you want to delete this time
t_{in}= D /
we substitute
t_{out} = \frac{v_s - v_r}{v_s+v_r} ()
t_{out} =