A rectangular metal loop with 0.053 ? resistance is placed next to one wire of t

Question

A rectangular metal loop with 0.053 ? resistance is placed next to one wire of the RC circuit shown in the figure below. The capacitor is charged to 20 V with the polarity shown, then the switch is closed at t = 0 s.

What is the current in the loop at t = 5.3 ?s? Assume that only the circuit wire next to the loop is close enough to produce a significant magnetic field. (Use 4? ? 10?7 T m/A for the permeability constant ?0.)

(I know the current is counterclockwise, by lenz's law since the flux through the loop is decreasing with time since the voltage source is a capacitor)

Explanation / Answer

Time constant of RC ckt, T = R*C

= 5*10^-6*2

= 10*10^-6 s

at t = 0, Imax = V/R

= 20/2

= 10 A

Current in the ckt at time t,

I = Imax*e^(-t/T)

dI/dt = Imax*e^(-t/T)*(1/T)

at t = 5.3 micro s,

dI/dt = (10/(10*10^-6))*e^(-5.3/10)

= 5.886*10^5 A/s

magntic flux through the loop = (mue*I*L/(2*pi))ln(1.5/0.5)

induced emf = the rate of change of magnetic flux through the loop

= (mue*L/(2*pi))ln(1.5/0.5)*dI/dt

= (4*pi*10^-7*0.02/(2*pi))*ln(1.5/0.5)*5.886*10^5

= 2.586*10^-3 volts

induced current = induced emf/R

= 2.586*10^-3/0.053

= 0.0488 A or 48.8 mA <<<<<<<<<--------------Answer

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