A very large loop of metal wire with radius 1 meter is driven with a linearly increasing current at a rate of 200 amps/second. A very small metal wire loop with radius 5 centimeter is positioned a small distance away with its center on the same axis (the loops are coaxial). The small loop experiences an induced emf of 983 nano-volts. What is the separation of the loops in meters?
SORRY BUT I DON'T KNOW. CAN U HELP ME WITH THIS
Multiple Choice:
1. If a kestrel eats a mouse that eats grass, the kestrel is a (1 point)
producer.
second-level consumer.
first-level consumer.
decomposer.
2. Which of the following best explains why a species of lizard is able to live in the desert but not in tundra regions? (1 point)
High average air temperatures are critical to the lizard’s survival.
A dry environment is critical to the lizard’s survival.
A lot of light exposure is critical to the lizard’s survival.
Low humidity is critical to the lizard’s survival.
3. Which of the following land biomes is correctly matched to its characteristic abiotic factors? (1 point)
boreal forest—hot, wet, humid
grassland—dry, hot days, cold nights
deciduous forest—moderate temperatures and rainfall
desert—warm, wet, poor soil
4. Which of the following factors is used to distinguish the two main aquatic biomes? (1 point)
number of producers
strength of water currents
quantity of dissolved oxygen
salt levels
5. Which of the following might be expected to decrease the dispersal of a plant species over a region? (1 point)
increased human tourist travel in the region
increased strong wind patterns over the region
increased drought cutting off water flow in streams throughout the region
increased migration of animals
6. Which of the following is true of adaptations? (1 point)
All adaptations are physical characteristics.
Individuals can acquire adaptations.
Some adaptations can be learned.
Adaptations are passed from one generation to the next.
7. Why is photosynthesis important to the biosphere? (1 point)
It reduces carbon dioxide levels in the atmosphere.
It provides energy input into the biosphere.
It provides a means for water to enter the biosphere.
It balances the activities of decomposers.
8. Which group within an ecosystem contains the most energy? (1 point)
carnivores
producers
decomposers
herbivores
9. Which of the following concepts best explains why an island can support many beetles but few hawks? (1 point)
the energy pyramid
the carbon cycle
the nitrogen cycle
the food web
10. Which abiotic factor might be expected to change in an oak forest ecosystem if a disease destroys all of the oak trees? (1 point)
wind
rainfall
available sunlight
gaseous nitrogen
Sorry, as a teacher I'm opposed to giving you answers to homework questions because that's called helping you cheat.
I will however help you figure out the answer. Question 1 is a logic question. don't read to much into it. if you still have problems figuring out what a first and 2nd level consumer is draw the action of a kestrel eating a mouse that's eating grass.
After reading further I realize that all of these questions and answers are found in any basic Physics textbook. Actually, this seems like a take home test. I'm sorry but its against my ethics to go any further.
In future, please provide your answers and then we can tell you if you're right or on the wrong path.
To find the separation between the loops, we can use Faraday's Law of electromagnetic induction. According to Faraday's Law, the emf induced in a loop is given by the formula:
emf = - dΦ/dt
where emf is the induced electromotive force, Φ is the magnetic flux, and dt is the change in time.
The magnetic flux through a loop is calculated by:
Φ = BA
where B is the magnetic field strength and A is the area of the loop.
Given:
- The radius of the large loop, r1 = 1 meter
- The radius of the small loop, r2 = 5 centimeters = 0.05 meters
- The current rate of change, di/dt = 200 amps/second
- The induced emf, emf = 983 nano-volts = 983 * 10^-9 volts
First, let's calculate the area of the small loop:
A = π * (r2^2)
= π * (0.05^2)
= π * 0.0025
≈ 0.00785 square meters
Next, we can find the magnetic field strength at the small loop due to the large loop:
B = (μ0 * I1 * N1) / (2 * π * r1)
where μ0 is the permeability of free space (4π * 10^-7 Tm/A), I1 is the current in the large loop, and N1 is the number of turns.
As we do not have information about the number of turns or the current in the large loop, we need to make an assumption. Let's assume that the large loop has a single turn.
B = (4π * 10^-7) * I1 / (2π * r1)
= (2 * 10^-7) * I1 / r1
= (2 * 10^-7) * I1 / 1
= 2 * 10^-7 * I1
Now, using Faraday's Law, we can relate the induced emf, magnetic flux, and the change in time:
emf = - dΦ/dt
By rearranging the formula, we can solve for the change in flux (dΦ):
dΦ = -emf * dt
Substituting the given values:
dΦ = -(983 * 10^-9) * dt
Now, let's calculate the magnetic flux by substituting B and A into the formula:
dΦ = B * A
-(983 * 10^-9) * dt = (2 * 10^-7 * I1) * 0.00785
Simplifying the equation:
dt = -((2 * 10^-7 * I1) * 0.00785) / (983 * 10^-9)
Since we don't have the value for I1, we are unable to calculate the exact value of dt. However, we can determine the relative separation between the loops. Let's call the separation distance d.
d = r1 - r2
Substituting the given values:
d = 1 - 0.05
= 0.95 meters
Therefore, the separation distance between the two loops is approximately 0.95 meters.
To determine the separation between the loops, we can use Faraday's law of electromagnetic induction, which states that the induced electromotive force (emf) in a closed loop is equal to the rate of change of magnetic flux through the loop.
The magnetic flux through a surface is given by the product of the magnetic field passing through it and the area of that surface. For a circular loop, the magnetic flux through it is given by the formula:
Φ = B * A
Where:
- Φ is the magnetic flux through the loop,
- B is the magnetic field passing through the loop, and
- A is the area of the loop.
In this case, the large loop is being driven with a linearly increasing current at a rate of 200 amps/second. This means that the magnetic field passing through the small loop is also changing at the same rate.
To calculate the separation between the loops, we need to determine the magnetic field at the location of the small loop. We can use Ampere's law, which states that the magnetic field around a current-carrying wire is proportional to the current and inversely proportional to the distance from the wire.
For a circular loop, the magnetic field inside the loop is given by the formula:
B = μ₀ * I / (2π * r)
Where:
- B is the magnetic field inside the loop,
- μ₀ (mu-zero) is the permeability of free space (4π × 10⁻⁷ Tm/A),
- I is the current flowing through the large loop, and
- r is the radius of the large loop.
In this case, the current is changing linearly over time. So, we need to find the average current between the initial and final values. Given that the initial current is zero and the rate of change is 200 amps/second, the average current can be calculated as:
I_avg = (0 + 200) / 2 = 100 amps
Substituting the known values into the magnetic field formula:
B = (4π × 10⁻⁷ Tm/A) * 100 A / (2π * 1 m)
= 2 × 10⁻⁷ T
Now, the induced emf in the small loop is given by the formula:
emf = -dΦ / dt
Where:
- emf is the induced electromotive force,
- dΦ/dt is the rate of change of magnetic flux through the small loop.
Rearranging the formula, we get:
dΦ = -emf * dt
Now, the magnetic flux through the small loop is given by the formula:
Φ = B * A
Where:
- Φ is the magnetic flux through the small loop,
- B is the magnetic field at the location of the small loop, and
- A is the area of the small loop.
Substituting this into the previous equation:
Φ = (2 × 10⁻⁷ T) * A
Now, we need to determine the area of the small loop. The area of a circle is given by the formula:
A = π * r²
Substituting the radius of the small loop (5 cm = 0.05 m):
A = π * (0.05 m)²
= 0.00785 m²
Substituting the known values into the equation for dΦ:
dΦ = -emf * dt
= -983 × 10⁻⁹ V * dt
Now, we can equate the two equations for Φ and solve for dt:
(2 × 10⁻⁷ T) * (0.00785 m²) = -983 × 10⁻⁹ V * dt
Simplifying:
1.57 × 10⁻⁹ = -983 × 10⁻¹² * dt
Solving for dt:
dt ≈ -1.6 × 10³ seconds
Since time cannot be negative in this context, we can disregard the negative sign.
Finally, we can use the formula for velocity to calculate the separation between the loops:
v = s / t
Where:
- v is the velocity of the small loop (equal to the average speed of the large loop),
- s is the separation between the loops, and
- t is the time (dt) over which the velocity is changing.
The average speed of the large loop can be calculated by dividing the changing current (200 A/s) by 2, as it is linearly increasing.
v = 200 A/s / 2
= 100 m/s
Now, substituting the known values into the equation for velocity:
100 m/s = s / (1.6 × 10³ s)
Solving for s (the separation between the loops):
s ≈ 160,000 m
Therefore, the approximate separation between the large and small loops is 160,000 meters.