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:
3A is the larger of the two currents.
Explanation:
Let the currents in the two wires be I₁ and I₂
given:
Magnitude of the electric field, B = 4.0μT = 4.0×10⁻⁶T
Distance, R = 10cm = 0.1m
Ratio of the current = I₁ : I₂ = 3 : 1
Now, the magnitude of a magnetic field at a distance 'R' due to the current 'I' is given as
Where is the magnitude constant = 4π×10⁻⁷ H/m
Thus, the magnitude of a magnetic field due to I₁ will be
given,
B = B₁ - B₂ (since both the currents are in the same direction and parallel)
substituting the values of B, B₁ and B₂
we get
4.0×10⁻⁶T = -
or
4.0×10⁻⁶T =
also
⇒
substituting the values in the above equation we get
4.0×10⁻⁶T =
⇒
also
⇒
⇒
Hence, the larger of the two currents is 3A
To solve this problem we will use the concepts related to angular motion equations. Therefore we will have that the angular acceleration will be equivalent to the change in the angular velocity per unit of time.
Later we will use the relationship between linear velocity, radius and angular velocity to find said angular velocity and use it in the mathematical expression of angular acceleration.
The average angular acceleration
Here
= Angular acceleration
Initial and final angular velocity
There is not initial angular velocity,then
We know that the relation between the tangential velocity with the angular velocity is given by,
Here,
r = Radius
= Angular velocity,
Rearranging to find the angular velocity
Remember that the radius is half te diameter.
Now replacing this expression at the first equation we have,
Therefore teh average angular acceleration of each wheel is