The Earth has an electric field with magnitude roughly 100 N/C at its surface.
A) Assuming there is a point charge at the earth's center creating this field, how much charge does the earth posses? Treat the earth as a sphere of radius 6.37 * 10^6 m.
B)As the earth orbits the sun,this charge moves and therefore creates a magnetic field. If we assume the earth moves on a circle ( whihc is approximately true) we can use the eqn for a current loop, B =(miu)I/2R, to find the magnetic field created by the earth( by the way, this is completely unrelated to the earth's North and south magnetic poles!) but first, find the current created by the earth as ut circles the sun.
HINT: I= Q/(delta)t. Think of a cross section of the earth' orbit. How often does a charge pass through it?
C) Using the earth sun distance , R= 1.5 * 10^11 m, calculate the magnetic field created by earth as it circles the sun.
A. E=kQ/r
B. The current earth creates is Q/timeofonerevolution. Ignore the hint, charges don't pass through the orbit.
charge dose (450.85)*10^3
A) To find the charge possessed by the Earth, we can use the formula for electric field strength due to a point charge:
E = k(Q/r^2)
where E is the electric field strength, k is the electrostatic constant (approximately 9 x 10^9 N m^2/C^2), Q is the charge, and r is the distance from the point charge.
Given that the electric field strength at the Earth's surface is 100 N/C, and the radius of the Earth is 6.37 x 10^6 m, we can rearrange the formula to solve for the charge:
Q = E * r^2 / k
Q = (100 N/C) * (6.37 x 10^6 m)^2 / (9 x 10^9 N m^2/C^2)
Calculating the above expression will give us the charge possessed by the Earth.
B) To find the current created by the Earth as it orbits the Sun, we need to determine the number of charges passing through a cross-section of the Earth's orbit per unit time.
The charge passing through the cross-section of the Earth's orbit is given by:
ΔQ = I * Δt
where ΔQ is the charge passing through, I is the current, and Δt is the change in time.
Since the Earth completes one orbit in one year (approximately 3.15 x 10^7 seconds), we can find the current using the rearranged equation:
I = ΔQ / Δt
I = Q / Δt
Remember, we found the charge possessed by the Earth in part A).
C) To calculate the magnetic field created by the Earth as it orbits the Sun, we can use the formula for the magnetic field due to a current loop:
B = (μ * I) / (2πR)
where B is the magnetic field, μ is the permeability of free space (approximately 4π x 10^-7 T m/A), I is the current, and R is the distance from the current loop.
Given that the distance from the Earth to the Sun is 1.5 x 10^11 m, we can plug in the values into the formula to calculate the magnetic field.
A) To find the amount of charge possessed by the Earth, we can use the formula for electric field intensity due to a point charge, E = k * (Q/r^2), where E is the electric field intensity, k is the electrostatic constant (approximately equal to 9 * 10^9 Nm^2/C^2), Q is the charge, and r is the distance from the point charge.
In this case, the electric field intensity is given as 100 N/C and the distance from the charge is the radius of the Earth, 6.37 * 10^6 m.
Plugging in these values into the formula, we can solve for Q:
100 N/C = (9 * 10^9 Nm^2/C^2) * (Q / (6.37 * 10^6 m)^2)
Simplifying the equation:
100 N/C = (9 * 10^9 Nm^2/C^2) * (Q / (4.07 * 10^13 m^2))
Divide both sides by (9 * 10^9 Nm^2/C^2):
(Q / (4.07 * 10^13 m^2)) = (100 N/C) / (9 * 10^9 Nm^2/C^2)
Simplify the right side:
(Q / (4.07 * 10^13 m^2)) = 1.11 * 10^-12 C
Multiply both sides by (4.07 * 10^13 m^2) to isolate Q:
Q = (1.11 * 10^-12 C) * (4.07 * 10^13 m^2)
Q ≈ 4.52 * 10^1 C
So, the Earth possesses approximately 45.2 C of charge.
B) To find the current created by the Earth as it circles the Sun, we can think of a cross-section of the Earth's orbit and calculate the charge passing through it over a given time interval.
The equation for current is I = Q / Δt, where I is the current, Q is the charge, and Δt is the time interval.
In a cross-section of the Earth's orbit, the charge passing through it can be considered as the charge of the Earth itself, which we found to be approximately 45.2 C.
To determine the time interval Δt, we need to consider the time it takes for the Earth to complete one orbit around the Sun. This is the orbital period of the Earth, which is approximately 365.25 days.
Converting the orbital period into seconds:
365.25 days * 24 hours/day * 60 minutes/hour * 60 seconds/minute ≈ 3.15576 * 10^7 seconds
So, the time interval Δt is approximately 3.15576 * 10^7 seconds.
Now, we can calculate the current I:
I = Q / Δt = (45.2 C) / (3.15576 * 10^7 s) ≈ 1.43 * 10^-6 A
Therefore, the current created by the Earth as it circles the Sun is approximately 1.43 μA (microamperes).
C) To calculate the magnetic field created by the Earth as it circles the Sun, we can use the equation B = (μ * I) / (2 * R), where B is the magnetic field, μ is the magnetic permeability of free space (approximately equal to 4π * 10^-7 Tm/A), I is the current, and R is the distance from the Sun.
Given that the Earth-Sun distance, R, is 1.5 * 10^11 m, and the current I is approximately 1.43 * 10^-6 A, we can plug these values into the equation:
B = (4π * 10^-7 Tm/A) * (1.43 * 10^-6 A) / (2 * (1.5 * 10^11 m))
Simplifying:
B = (5.655 * 10^-7 Tm) / (3 * 10^11 m)
B ≈ 1.885 * 10^-18 T
Therefore, the magnetic field created by the Earth as it circles the Sun is approximately 1.885 attoteslas.