Please help!! I have tried everything and is still coming up with and incorrect answer.
A piece of copper wire is formed into a single circular loop of radius 13 cm. A magnetic field is oriented parallel to the normal to the loop, and it increases from 0 to 0.70 T in a time of 0.45 s. The wire has a resistance per unit length of 3.3 10-2 /m. What is the average electrical energy dissipated in the resistance of the wire.
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Physic please help! - drwls, Tuesday, October 9, 2007 at 7:49pm
The change in B divided by the time (0.45 s), multiplied by the loop area in square meters, is the average voltage in the loop during the interval. The wire resistance is 2 pi R * 3.3 10-2 ohms(?)/m.
Coumpute the average current (Vav/R) and multiply it by the average voltage, Vav, computed by the method above.
Average Power =
[(delta B)(pi R^2)/(delta t)]^2/(2 pi R)
= (1/2) pi* R^3 [(delta B)/(delta t)]^2
Check my math and my logic. I could be wrong
To solve this problem, we need to use the formula for power:
Average Power = (Current^2) * Resistance [P = I^2 * R]
Now, let's break down the problem and find the values we need for the formula:
1. We are given the loop radius, which is 13 cm. Convert it to meters by dividing it by 100: R = 0.13 m.
2. The magnetic field, B, increases from 0 to 0.70 T in a time of 0.45 s. In other words, we have a change in magnetic field, ΔB = 0.70 T - 0 T = 0.70 T. And the change in time is Δt = 0.45 s.
3. We are given the resistance per unit length of the wire, which is 3.3 * 10^(-2) Ω/m.
Now, let's solve for the average power dissipated in the resistance of the wire:
Step 1: Calculate the average voltage (Vav):
Using the formula given by drwls, Vav = [(ΔB)(πR^2)/(Δt)].
Plugging in the values we know:
Vav = [(0.70 T)(π * (0.13 m)^2)/(0.45 s)].
Step 2: Calculate the current (I):
Using Ohm's Law, we can find the current I = Vav/R, where R is the resistance per unit length of the wire.
Plugging in the values we know:
I = Vav / (2πR) = [(0.70 T)(π * (0.13 m)^2)/(0.45 s)] / (2π(3.3 * 10^(-2) Ω/m)).
Step 3: Calculate the average power (P) using the formula:
P = I^2 * R.
Plugging in the values we know:
P = I^2 * (2πR) = [(0.70 T)(π * (0.13 m)^2)/(0.45 s)]^2 * (2π(3.3*10^(-2) Ω/m)).
Simplify the equation and calculate the final answer using a calculator.
Please note that the final answer will be in watts (W), which is the unit of power.
To solve this problem, we need to follow the steps provided by drwls. Let's break it down step by step:
Step 1: Calculate the change in magnetic field, ΔB.
ΔB = final magnetic field - initial magnetic field
Given: final magnetic field = 0.70 T, initial magnetic field = 0 T
ΔB = 0.70 T - 0 T
ΔB = 0.70 T
Step 2: Calculate the loop area, A.
Given: radius of the loop = 13 cm = 0.13 m
A = π * (radius)^2
A = π * (0.13 m)^2
A = 0.0534 m^2
Step 3: Calculate the average voltage in the loop, Vav.
Vav = (ΔB / Δt) * A
Given: Δt = 0.45 s
Vav = (0.70 T / 0.45 s) * 0.0534 m^2
Vav = 0.1098 V
Step 4: Calculate the wire resistance, R.
Given: resistance per unit length = 3.3 * 10^-2 Ω/m
Since we are dealing with a circular loop, the length of the wire is equal to the circumference of the loop, which is 2π times the radius.
R = resistance per unit length * circumference
Given: radius of the loop = 0.13 m
Circumference = 2π * radius
Circumference = 2π * 0.13 m
R = 3.3 * 10^-2 Ω/m * (2π * 0.13 m)
R = 0.085 Ω
Step 5: Calculate the average current, Iavg.
Iavg = Vav / R
Iavg = 0.1098 V / 0.085 Ω
Iavg = 1.2906 A
Step 6: Calculate the average electrical energy dissipated in the resistance, Pavg.
Pavg = (Iavg^2) * R
Pavg = (1.2906 A)^2 * 0.085 Ω
Pavg = 0.1489 W
Therefore, the average electrical energy dissipated in the resistance of the wire is approximately 0.1489 W.