in the diagram below the outer wire has a resistance of 6 and the circular part
ID: 1537714 • Letter: I
Question
in the diagram below the outer wire has a resistance of 6 and the circular part of the wire has a radius of 15 cm and the battery is a 120 V battery. A) when the switch is first closed, the current in the outer loop goes from 0 to its maximum in 25 ms. What is the induced emf as well as the induced current in the inner loop of wire during this time if it's radius is 8 cm and it's resistance is .03 ohms? B) if the inner loop was replaced with another loop of wire that carried the same current as the induced current found above, what would the radius of the new inner loop need to be so that the magnetic field at the center was zero? has a total resistance of battery is a 120 V battery. in the outer loop goes from as the induced current in t mintance is 0030Q?Explanation / Answer
Resistance of outer wire = R = 6 ohms
circular part radius= r = 0.15 m
Battery EMF, V = 120 V
A) initial current = 0
final current i = V/R = 120/6 = 20 A
masgnetic field of a current carrying loop, is B = mu*I/2r
final flux = mu*i*pi*r'^2/2r [ r' is the radiua od the inner ring]
flux = 4*pi*10^-7*20*3.14*0.08^2/2*0.15 = 8.4135*10^-7
rate of flux chanfe = flux/time = 3.365*10^-5 V
induced emf = 3.365*10^-5 V
induced i = v/R' = 3.365*10^-5/0.03 = 1.12 mA
B) this induced current has B at centre = mu*i/2R' = mu*I/2R
R' = ir/I = 8.4*10^-3 mm
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