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 answer

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: \( \approx 0.416 , V \)