A circular wire loop carrying an alternating current I(t)=I0sin(2?ft) lies flat

Question

A circular wire loop carrying an alternating current I(t)=I0sin(2?ft) lies flat in the x-y plane centered at the origin (radius of the loop r=1.0 m, current amplitude I0=1.0 A, cycle frequency f=60 Hz).

(A) Find the magnetic field B(t) at the center of the loop and express it as a function of time.

A smaller circular loop with radius a=0.01 m is placed at the origin parallel to the big loop. The resistance of the small loop is R=5.0 Ohms.

(B)What is the magnetic flux passing through the small loop as a function of time?

(C) Find the emf induced in the small loop and express it as a function of time.

(D)Find the current in the small loop as a function of time. What is the ratio of the peak current in the small loop to the peak current in the big loop? Which of the quantities given in this problem does the ratio depend on?

Explanation / Answer

a) B = u0 I/ 2 r = 4*pi*10^(-7)*1/(2*1) * sin(2*pi*60*t)=6.28E-7*sin(120 pi t)

B) flux = B A = 6.28E-7*sin(120 pi t)*pi*0.01^2=6.28E-11*sin(120*pi*t)

c) emf = dflux/dt = 6.28E-11*120*pi*cos(120*pi*t)=2.37E-8*cos(120*pi*t)

d) current = emf/R = 4.74E-9*cos(120*pi*t)

d) Ismall/Ibig = 4.74E-9/1 = 4.74E-9, depends on f r and a and R

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