A square-wave voltage of 100V with frequency of 60Hz is applied on a coilwound on a closed iron core. The coil has 500 turns, and the cross-section areaof the core is 0.001m2. If the core has no resistance, determinea) the maximum value of flux;b) the maximum value of voltage with the maximum flux density of 1.2T
Question
A square-wave voltage of 100V with frequency of 60Hz is applied on a coilwound on a closed iron core. The coil has 500 turns, and the cross-section areaof the core is 0.001m2. If the core has no resistance, determinea) the maximum value of flux;b) the maximum value of voltage with the maximum flux density of 1.2T
Solution
a) The maximum value of flux:
The maximum flux (Φmax) in a magnetic field is given by the formula:
Φmax = Bmax * A
Where: Bmax is the maximum magnetic field strength (also known as magnetic flux density), and A is the cross-sectional area of the coil.
However, in this case, we are not given Bmax directly. Instead, we are given the voltage (V) and the number of turns in the coil (N). We can use Faraday's law of electromagnetic induction to find Bmax. Faraday's law states that the induced voltage in a coil is equal to the rate of change of magnetic flux through the coil. In the case of a square wave, the rate of change of flux is at its maximum when the voltage is at its maximum. Therefore, we can write:
V = N * d(Φ)/dt
Where: d(Φ)/dt is the rate of change of flux.
Rearranging for d(Φ)/dt gives:
d(Φ)/dt = V / N
Since the rate of change of flux is at its maximum when the voltage is at its maximum, we can write:
d(Φmax)/dt = Vmax / N
Rearranging for Φmax gives:
Φmax = (Vmax / N) * t
Where: t is the period of one cycle of the square wave.
The frequency (f) of the square wave is given as 60Hz, so:
t = 1/f = 1/60 = 0.0167s
Substituting the given values into the equation gives:
Φmax = (100V / 500 turns) * 0.0167s = 0.00334 Wb
b) The maximum value of voltage with the maximum flux density of 1.2T:
We can use the formula for the maximum flux (Φmax = Bmax * A) to find the maximum voltage. Rearranging for Bmax gives:
Bmax = Φmax / A
Substituting the given values into the equation gives:
Bmax = 0.00334 Wb / 0.001m^2 = 3.34T
However, the problem states that the maximum flux density is 1.2T. Therefore, the actual maximum flux density (Bmax_actual) is:
Bmax_actual = 1.2T
The maximum voltage (Vmax) is then given by:
Vmax = N * d(Φ)/dt
Substituting the given values into the equation gives:
Vmax = 500 turns * (1.2T * 0.001m^2) / 0.0167s = 35.93V
Therefore, the maximum value of voltage with the maximum flux density of 1.2T is approximately 36V.
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