Parts C and D please Class Management Help EXPTA 09 Begin Date: 3/13/2018 12:01:
ID: 777110 • Letter: P
Question
Parts C and D please
Class Management Help EXPTA 09 Begin Date: 3/13/2018 12:01:00 AM -- Due Date: 3/16/2018 11:59:00 PM End Date: 3/20/2018 11:59:00 PM (11%) Problem S: A square loop of side length a =2.2 cm is placed a distance b = 1.2 cm from a long wire carrying a current that varies with time at a constant rate, i.e lit) = Ot where is a constant Q =3.6 As. Randomized Variables a= 2.2 cm b= 1.2 cm Q = 3.6 As 25% Part (a) Use Ampere's law to find the magnetic field due to the wire as a function of time t at a distance r from the wire, in terms of a, b, Q, t, r, and fundamental constants 25% Part (b) What is the flux through the loop? Select the correct expression. Final Grade 100% Student Answer Correct Answer Grade Detail Student Feedback Correct Final Answer Credit 100% Submission Historv Answer Totals 0% 0% Hints Feedback 0% 0% 0% 0% 0% 0% Totals 25% Part (c) If the loop has a resistance of 2.5 , how much induced current flows in the loop? 25% Part (d) In what direction does this current flow? -Explanation / Answer
induced emf e = rate of change of magnetic flux
emf e = uo Q at/2pi * ln (1 + a/b) /dt
emf e = uo at/2pi * ln (1 +a/b) * dQ/dt
emf e = 4pi*10^-7 * 3.6*0.022/(2pi) * ln(1 + 2.2/1.2)
emf e = 0.165*10^-7 Volts
Induced current I = emf/R
I = 0.165*10^-7/(2.5)
I = 6.6 nano AMps
------------------------
Current Flows in CCW direction