1. -1 points SerPSE9 31.P001 My Notes Ask Your A flat loop of wire consisting of
ID: 2305312 • Letter: 1
Question
1. -1 points SerPSE9 31.P001 My Notes Ask Your A flat loop of wire consisting of a single turn of cross sectional area 8.20 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 1.60 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 1.80 ?? mA Need Help? Readt 2. -1 polnts SerPSE9 31.P.004.WI My Notes Ask Your A 19-tum circular coll of wire has diameter 1.08 m. It is placed with its axis along the direction or the Earth's ma?netic fleld of 59.0 ?? and then in 0.200 $ ls nipped 180°. An average emt or what magnitude Is generated In the coll? mv 3. -1 pointa SerPSE9 31.P.005. My Note3 AgkYour The flexible loop in the figure below has a radius of 11.0 cm and is in a magnetic field of magnitude 0.100 T. The loop is grasped at points A and B and stretched until its area is nearly zero. If it takes 0.210 s to close the loop, what is the magnitude of the average induced emf in it during this time interval? mv Need HelpReadItExplanation / Answer
1. induced emf = rate of change of flux
= A dB/dt
= (8.20 x 10^-4 m^2) ( (1.60 - 0.5/1)
= 9.02 x 10^-4 Volt
I = e / R = 5.0 x 10^-4 A OR 0.50 mA
Ans: 0.50
2. induced emf = d(magnetic flux)/dt
= (NBA - (- N B A)) / t
= 2 N B A / t
= (2 x 19 x 59 x 10^-6)(pi (1.08/2)^2) / (0.200)
= 0.0103 V Or 10.3 mA
Ans: 10.3
3. induced emf = B dA/dt
= (0.10) ((pi 0.11^2) - 0) /(0.210)
= 0.0181 V Or 18.1 mV
Ans: 18.1
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.