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. When the wire's current rises from 5 A to 10 A, the strength of its magnetic field doubles
Explanation: The power of the magnetic field produced by a wire with a current flowing through it can be expressed as:
where signifies the permeability of free space, I denotes the current through the wire, and r represents the distance from it.
Observing the equation, we find that the magnetic field's intensity directly increases with the current: thus when the current goes from 5 A to 10 A, it is effectively doubling the magnetic field as well.
2. When the distance from the wire to the charged particle changes from 10 cm to 20 cm, the strength of the magnetic field reduces by half
Explanation: The intensity of the magnetic field created by a wire carrying a current is also given as:
This shows that the magnetic field decreases proportionally with the distance from the wire (r). In this scenario, as the distance from 10 cm to 20 cm is doubled, the result is that the field's strength will drop to half the original value.
3. The direction of the force changes
Explanation: The force experienced by a charged particle within a magnetic field is calculated by:
where q is the charge, v is the particle's velocity, B indicates the strength of the magnetic field, and defines the angle between v and B's direction. If the particle's charge switches from 2 µC to –2 µC, the magnitude of the force remains stable (due to the absolute value of q being unchanged), but since q gains a negative sign (-), the force's sign also flips, resulting in the force changing direction.
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The alteration in potential energy is 
In the query, it is stated that
The intensity of the uniform electric field equals The distance the electron covers is
Typically, the force exerted on this electron is expressed mathematically as
Where F signifies the force and q represents the charge of the electron, which is a fixed value of
Thus
Generally, the work-energy theorem is mathematically framed as
Where W denotes the work done on the electron by the electric field and is the change in kinetic energy
Additionally, work done on the electron can also be described as
Where assuming that the electron's movement aligns with the x-axis
So
Inserting values
According to the conservation of energy
Where signifies the change in potential energy
Thus
Response:
Clarification:
Refer to the diagram indicating the charges on the specified sphere (see attachment).
The electric field at the stated positions is
E(r) = 0 for r≤a. Equation 1
E(r) = kq/r² for a<r<b. Equation 2
E(r) = 0 for b<r<c. Equation 3
E(r) = kq/r² for r>c. Equation 4.
We understand that electric potential correlates with the electric field through
V = Ed
A. To compute the potential at the outer surface of the hollow sphere (r=c), we determine that the electric field there is
E = kQ / r²
Then,
V = Ed,
At d = r = c
Thus,
Vc = (kQ / c²) × c
Vc = kQ / c
As a result, the total charge Q consists of +q, -q, and +q
Hence, Q = q - q + q = q
V = kq / c
B. To calculate the potential at the inner surface of the hollow sphere (r=b), we have
V = kQ/r
V = kQ / b, noting that r = b
So, Q = q
V = kq / b
C. At r = a
Following from equation 1:
E(r) = 0 for r≤a. Equation 1
The electric field at the surface of the solid sphere is 0, E = 0N/C
Thus,
V = Ed = 0 V
Consequently, the electric potential at the solid sphere's surface is 0.
D. At r = 0
The electric potential can be determined by
V = kq / r
As r approaches 0,
V = kq / 0
V approaches infinity.
