Question
A coil is wound with 1200 turns of copper wire that has a cross-sectional area of 2,5mm² and a mean length per turn of 25cm. If the coil has an inductance of 48×10-⁶Henry's, calculate:
1) the resistance of the winding if the specific resistance of the wire is 0,017×10-⁶ ohm/ meter
2) the average value of the emf induced in the coil when a current of 6,5A is reversed in 1,5ms.
1) the resistance of the winding if the specific resistance of the wire is 0,017×10-⁶ ohm/ meter
2) the average value of the emf induced in the coil when a current of 6,5A is reversed in 1,5ms.
Answers
GPT-4o mini
To solve the problem, we will perform two calculations: the resistance of the winding and the average value of the induced emf when the current is reversed.
### 1) Calculate the resistance of the winding
The resistance \( R \) of the coil can be calculated using the formula:
\[
R = \rho \frac{L}{A}
\]
where:
- \( \rho \) = specific resistance of the copper wire = \( 0.017 \times 10^{-6} \, \Omega \cdot m \)
- \( L \) = total length of the wire (in meters)
- \( A \) = cross-sectional area of the wire (in \( m^2 \))
First, we need to find the total length of the wire. The mean length per turn is given as 25 cm (which is \( 0.25 \, m \)), and the number of turns is 1200.
Thus, the total length \( L \) is:
\[
L = \text{number of turns} \times \text{mean length per turn} = 1200 \times 0.25 \, m = 300 \, m
\]
Next, convert the cross-sectional area from \( mm^2 \) to \( m^2 \):
\[
A = 2.5 \, mm^2 = 2.5 \times 10^{-6} \, m^2
\]
Now we can substitute into the resistance formula:
\[
R = 0.017 \times 10^{-6} \cdot \frac{300}{2.5 \times 10^{-6}}
\]
Calculating:
\[
R = 0.017 \times 10^{-6} \cdot \frac{300}{2.5 \times 10^{-6}} = 0.017 \cdot \frac{300}{2.5} = 0.017 \cdot 120 = 2.04 \times 10^{-2} \, \Omega
\]
Thus, the resistance of the winding is:
\[
R \approx 0.0204 \, \Omega
\]
### 2) Calculate the average value of the emf induced in the coil
The emf induced in the coil can be calculated using Faraday’s law of electromagnetic induction, given by:
\[
\text{emf} = -L \frac{\Delta I}{\Delta t}
\]
where:
- \( L \) = inductance of the coil = \( 48 \times 10^{-6} \, H \)
- \( \Delta I \) = change in current = \( I_{final} - I_{initial} \) (Since the current is reversed, we take the change from \( +6.5 \, A \) to \( -6.5 \, A \))
- \( \Delta t \) = time duration of the change = \( 1.5 \, ms = 1.5 \times 10^{-3} \, s \)
Calculating \( \Delta I \):
\[
\Delta I = -6.5 - 6.5 = -13 \, A
\]
Now substitute the values into the emf formula:
\[
\text{emf} = -48 \times 10^{-6} \cdot \frac{-13}{1.5 \times 10^{-3}}
\]
Calculating:
\[
\text{emf} = 48 \times 10^{-6} \cdot \frac{13}{1.5 \times 10^{-3}} = 48 \times 10^{-6} \cdot \frac{13 \times 10^3}{1.5} = 48 \times 10^{-6} \cdot 8666.67
\]
\[
\text{emf} = 0.416 \, V
\]
Thus, the average value of the emf induced in the coil is approximately:
\[
\text{emf} \approx 0.416 \, V
\]
### Summary of Results
1) The resistance of the winding: \( \approx 0.0204 \, \Omega \)
2) The average value of the emf induced: \( \approx 0.416 \, V \)
### 1) Calculate the resistance of the winding
The resistance \( R \) of the coil can be calculated using the formula:
\[
R = \rho \frac{L}{A}
\]
where:
- \( \rho \) = specific resistance of the copper wire = \( 0.017 \times 10^{-6} \, \Omega \cdot m \)
- \( L \) = total length of the wire (in meters)
- \( A \) = cross-sectional area of the wire (in \( m^2 \))
First, we need to find the total length of the wire. The mean length per turn is given as 25 cm (which is \( 0.25 \, m \)), and the number of turns is 1200.
Thus, the total length \( L \) is:
\[
L = \text{number of turns} \times \text{mean length per turn} = 1200 \times 0.25 \, m = 300 \, m
\]
Next, convert the cross-sectional area from \( mm^2 \) to \( m^2 \):
\[
A = 2.5 \, mm^2 = 2.5 \times 10^{-6} \, m^2
\]
Now we can substitute into the resistance formula:
\[
R = 0.017 \times 10^{-6} \cdot \frac{300}{2.5 \times 10^{-6}}
\]
Calculating:
\[
R = 0.017 \times 10^{-6} \cdot \frac{300}{2.5 \times 10^{-6}} = 0.017 \cdot \frac{300}{2.5} = 0.017 \cdot 120 = 2.04 \times 10^{-2} \, \Omega
\]
Thus, the resistance of the winding is:
\[
R \approx 0.0204 \, \Omega
\]
### 2) Calculate the average value of the emf induced in the coil
The emf induced in the coil can be calculated using Faraday’s law of electromagnetic induction, given by:
\[
\text{emf} = -L \frac{\Delta I}{\Delta t}
\]
where:
- \( L \) = inductance of the coil = \( 48 \times 10^{-6} \, H \)
- \( \Delta I \) = change in current = \( I_{final} - I_{initial} \) (Since the current is reversed, we take the change from \( +6.5 \, A \) to \( -6.5 \, A \))
- \( \Delta t \) = time duration of the change = \( 1.5 \, ms = 1.5 \times 10^{-3} \, s \)
Calculating \( \Delta I \):
\[
\Delta I = -6.5 - 6.5 = -13 \, A
\]
Now substitute the values into the emf formula:
\[
\text{emf} = -48 \times 10^{-6} \cdot \frac{-13}{1.5 \times 10^{-3}}
\]
Calculating:
\[
\text{emf} = 48 \times 10^{-6} \cdot \frac{13}{1.5 \times 10^{-3}} = 48 \times 10^{-6} \cdot \frac{13 \times 10^3}{1.5} = 48 \times 10^{-6} \cdot 8666.67
\]
\[
\text{emf} = 0.416 \, V
\]
Thus, the average value of the emf induced in the coil is approximately:
\[
\text{emf} \approx 0.416 \, V
\]
### Summary of Results
1) The resistance of the winding: \( \approx 0.0204 \, \Omega \)
2) The average value of the emf induced: \( \approx 0.416 \, V \)