What happens to the current in a circuit if a 1.5 volt battery
1 answer:
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Answer:
B = 38.2μT
Explanation:
By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:
(1)
μo: magnetic permeability of vacuum = 4π*10^-7 T/A
r: distance from the center of the cylinder, in which B is calculated
Ir: current for the distance r
In this case, you first calculate the current Ir, by using the following relation:
J: current density
Ar: cross sectional area for r in the hollow cylinder
Ar is given by
The current density is given by the total area and the total current:
R2: outer radius = 26mm = 26*10^-3 m
R1: inner radius = 5 mm = 5*10^-3 m
IT: total current = 4 A
Then, the current in the wire for a distance r is:
(2)
You replace the last result of equation (2) into the equation (1):
Finally. you replace the values of all parameters:
hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT
Answer:
The ball traveled 0.827 m
Explanation:
Given;
distance between the metal plates of the room, d = 3.1 m
mass of the glass, m = 1.1g
charge on the glass, q = 4.7 nC
speed of the glass ball, v = 4.8 m/s
voltage of the ceiling, V = +3.0 x 10⁶ V
The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;
F = qV/d
|F| = (4.7 x 10⁻⁹ x 3 x 10⁶) / (3.1)
|F| = 4.548 x 10⁻³ N
F = - 4.548 x 10⁻³ N
The net horizontal force experienced by this ball is;
The work done between the ends of the plate is equal to product of the magnitude of net force on the ball and the distance traveled by the ball.
W = K.E
Therefore, the ball traveled 0.827 m