Question

1. A generator has a 150-turn square coil, 5cm on a side.

Part A

Find the peak emf in this coil when it's rotating at 110Hz in Earth's field, B=5

Answer 1

1.

a)

angular frequency

W=2pi*f =2pi*110=691.15 rad/s

Area

A=S²=0.05²=2.5*10^-3 m²

Peak Emf in the coil is

E_peak=NBAW=150*(5*10^-5)(2.5*10^-3)(691.15)

E_peak=0.01296 Volts =0.013 Volts (approx)

b)

E_peak=NBAW=150*(6.2)(2.5*10^-3)(691.15)

E_peak=1606.92 Volts

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2

Power loss in transmission line

P_input=V²/R =(12*10³)²/0.14

P_input=1.03*10⁹ Watts

Fraction of power lost in transmission line is

P/P_input =1.7*10³/(1.03*10⁹)=1.65*10^-6

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3.

Inductance

L=N*phi/I

L=240*0.85/(310*10^-3)

L=658.06 H

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4.

Energy stored in inductance

E=(1/2)LI²

=>L=2*E/I² =2*1.1/9²

L=0.0272 H

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5.

Resonsant frequecy

f=1/2pi*sqrt[LC)

So inductance

=>L=1/4pi²f²C

at f=94 MHz=>C=8.7 pF

L=1/4pi²*(94*10⁶)²*(8.7*10^-12)

L=3.3*10^-7 H