A rectangular metal loop with 0.050 W resistance is placed next to one wire of t

Question

A rectangular metal loop with 0.050 W resistance is placed next to one wire of the RC circuit shown below. The capacitor is charged to 20 V with the polarity shown, then the switch is closed at t=0 s. The closest edge of the loop is 0.50 cm from the circuit, and the loop is 1.0 cm wide and 2.0 cm long.

a) What is the direction of current in the loop for t>0 s? (draw on the diagram)

b) What is the current in the loop at t=5.0 ms? Assume that only the circuit wire next to the loop is close enough to produce a significant magnetic field.

Seeing all of the steps nice and clear would be super helpful. Trying to actually understand it all and not just get the answer.

Explanation / Answer

a) current direction is from positive plate to negative plate
b)Given that the potential difference is V=20V
the capacitance is C=5 x 10-6 F
resistance is R=2?
then the current at t=0 is
i0=V/R
=10A
the time constant is T=RC=10-5 s
then the current at time t= 5 x 10-6 sec is
i=i0 x e-t/T
=10 e -0.5
=6.06A
the area of the loop is A=2 x 10-4 m 2   
magnetic field at a distance 0.5cm is
B1=?i/2?d1
=2.424 x 10-4
magnetic field at a distance 1.5cm is
B2=?i/2?d2
=0.808 x 10-4
so the change in magnetic field is dB=1.616x 10-4
hence the induced emf in the circuit is E=A dB/T
=2 x 10-4 x 1.616 x 10^-4/10^-5
=3.232 x 10^-5
hence the current in the loop is I=
E/R
=3.232 x 10^-5/0.050
=0.53mA

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