A coil having :

N= 440 turns
diameter =1,2mm
Mean length= 120mm
Inductance= 400micro Henry's
Calculate the resistance is the resistivity of the wire is 0,027micro-ohm meters
Calculate the average value of the EMF induced in a coil when current of 4A is reversed in 0,001 seconds

1 answer

To solve your problem, we'll tackle it step by step.

Step 1: Calculate the Resistance

To calculate the resistance \( R \) of the coil, we can use the formula:

\[ R = \rho \frac{L}{A} \]

where:

  • \( R \) is the resistance,
  • \( \rho \) is the resistivity of the wire,
  • \( L \) is the length of the wire,
  • \( A \) is the cross-sectional area of the wire.

Given values:

  • \( \rho = 0.027 , \mu\Omega , m = 0.027 \times 10^{-6} , \Omega, m \)
  • Diameter of the wire = 1.2 mm = 0.0012 m
  • Mean length \( L = 120 , mm = 0.12 , m \)
  • \( N = 440 , turns \)

Calculate \( A \):

The cross-sectional area \( A \) of the wire can be calculated using the formula for the area of a circle:

\[ A = \pi \left(\frac{d}{2}\right)^2 = \pi \left(\frac{0.0012}{2}\right)^2 \]

Calculating it:

\[ A = \pi \left(0.0006\right)^2 \approx \pi \times 0.00000036 \approx 1.131 \times 10^{-6} , m^2 \]

Calculate \( R \):

Now substituting \( \rho \), \( L \), and \( A \) into the resistance formula:

\[ R = 0.027 \times 10^{-6} \frac{0.12}{1.131 \times 10^{-6}} \approx 0.027 \times 10^{-6} \times 106,272.5 \]

Calculating it:

\[ R \approx 0.027 \times 0.000028707 \approx 0.000000773 , \Omega = 0.773 , \mu\Omega \]

Step 2: Calculate the Average EMF Induced

The average EMF \( \mathcal{E} \) induced in the coil can be calculated using Faraday's law of electromagnetic induction:

\[ \mathcal{E} = -L \frac{\Delta I}{\Delta t} \]

Where:

  • \( L \) is the inductance (in henries),
  • \( \Delta I \) is the change in current,
  • \( \Delta t \) is the change in time.

Given values:

  • \( L = 400 , \mu H = 400 \times 10^{-6} , H \)
  • Change in current \( \Delta I = 4A - (-4A) = 4A + 4A = 8A \)
  • Change in time \( \Delta t = 0.001, s \)

Calculate \( \mathcal{E} \):

Substituting the values:

\[ \mathcal{E} = -400 \times 10^{-6} \frac{8}{0.001} \]

Calculating it:

\[ \mathcal{E} = -400 \times 10^{-6} \times 8000 = -3.2 , V \]

Thus, the average value of the EMF induced in the coil is approximately 3.2 V (the negative sign indicates the direction of induced EMF opposing the change in current).

Final Results:

  • Resistance of the coil: \( R \approx 0.773 , \mu\Omega \)
  • Average EMF induced: \( |\mathcal{E}| \approx 3.2 , V \)