1) Current in the wire: 0.0875 A
The current in the wire is given by:
where
Q is the charge passing a given point in the conductor
t is the time elapsed
In this problem, we have
Q = 420 C is the total charge passing through a given point in a time of
t = 80 min = 4800 s
So, the current is
2) Drift velocity of the electrons: 
The drift velocity of the electrons in the wire is given by:
where
I = 0.0875 A is the current
is the number of free electrons per cubic meter
A is the cross-sectional area
is the charge of one electron
The radius of the wire is
So the cross-sectional area is
So, the drift velocity is
Prior to touching the bar magnet, the magnetic domains in the nail were pointing in random directions. When Taylor touched the nail to the bar magnet the magnetic fields of the magnetic domains aligned and it became a temporary magnet.
Answer: 3 x 10^-24 watt
Explanation:
P ( resistivity) = 1.72e-8 (from the chart).
L= 2pi r
r= 30 cm.
R= pL/A
A= pi* r1^2
r1= 0.8118/2 * 10^-3 m
R= 1.68 x 10^-8 x (2x3.142x0.3)
= 3.24 x 10^-8
E=N do/dt
do= B* A
A= pi* 0.3^2
N=1
E = 1 x (14 x 3.142x 0.09) = 3.95
I=v/R
v=E,
I = 3.95 / 3.24 x 10^-8 = 1.22 x 10^8
P=I^2 x R.
= 3 x 10^-24 watt
Explanation:
Initial time, t₁ = 2:30 pm
Final time, t₂ = 2:30:45
We need to find the motion of students in terms of time. Final time is 45 seconds more than the initial time.
Change in time,
Hence, this is the required solution.
Models show how the atoms in a compound are connected.
