Answer:
a)n= 3.125 x electrons.
b)J= 1.515 x A/m²
c) =1.114 x m/s
d) see explanation
Explanation:
Current 'I' = 5A =>5C/s
diameter 'd'= 2.05 x m
radius 'r' = d/2 => 1.025 x m
no. of electrons 'n'= 8.5 x
a) the amount of electrons pass through the light bulb each second can be determined by:
I= Q/t
Q= I x t => 5 x 1
Q= 5C
As we know that: Q= ne
where e is the charge of electron i.e 1.6 x C
n= Q/e => 5/ 1.6 x
n= 3.125 x electrons.
b) the current density 'J' in the wire is given by
J= I/A => I/πr²
J= 5 / (3.14 x (1.025x )²)
J= 1.515 x A/m²
c) The typical speed'' of an electron is given by:
=
=1.515 x / 8.5 x x |-1.6 x |
=1.114 x m/s
d) According to these equations,
J= I/A
= =
If you were to use wire of twice the diameter, the current density and drift speed will change
Increase in the diameter increase the cross sectional area and decreases the current density as it has inverse relation.
Also drift velocity will decrease as it is inversely proportional to the area
To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
At time , Induced emf is,
Therefore the magnitude of the induced emf is 10.9V
Work equals force × displacement (distance between initial point and end point is displacement)
if u follow this it becomes
work = 50 × 2 which is equal to 100
comment if u have more questions
As thermal energy increases, there is more particle movement. As thermal energy increases, there is more particle movement. As thermal energy increases, there is less particle movement.
Sure hope this helps you
Answer:
<u>Electromagnetic introduction</u> is the production of an electromotive force (voltage) across an electrical conductor in a changing magnetic field.
- <em><u>Step up transformers</u></em><u> is</u> a transformer in which the output (secondary) voltage is greater than its input (primary) voltage is called a step-up transformer. The step-up transformer decreases the output current for keeping the input and output power of the system equal.
- <u><em>Step down transformer is </em></u><em>a transformer in which the output (secondary) voltage is less than its input (primary) voltage is called a step-down transformer. The number of turns on the primary of the transformer is greater than the turn on the secondary of the transformer.</em>
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<u>The difference between them:</u>
A transformer is a static device which transfers a.c electrical power from one circuit to the other at the same frequency, but the voltage level is usually changed. For economical reasons, electric power is required to be transmitted at high voltage whereas it has to be utilized at low voltage from a safety point of view. This increase in voltage for transmission and decrease in voltage for utilization can only be achieved by using a step-up and step-down transformer.
Hopefully this helped.