I solved part a but can someone go through the whole thing. ESPECIALLY part B an

Question

I solved part a but can someone go through the whole thing. ESPECIALLY part B and please do it Legibally and show all work!! Problem 19: (a) The area of an elastic circular loop decreases at a constant rate dA/dt 3.50 x10 m /s. The loop is in a magnetic field B 0.28 T whose direction is perpendicular to the plane of the loop. At t-0, the has area A 0.285 m Determine the loop emf at t 0, and at 200s. (b) Suppose the radius of the elastic loop in part (a) increases at a constant rate, dridt 4.30 cm/s. Determine the emf induced in the loop at t o s. (a) 9.8 mV (b) 23 mv 26 mV da so

Explanation / Answer

part b:

induced emf=-rate of change of flux

flux=magnetic field*area

=B*A

=B*pi*r^2

so induced emf=-d(B*pi*r^2)/dt

=-pi*B*2*r*(dr/dt)

given B=0.28 T

at t=0, r=sqrt(A/pi)

=0.30119 m

dr/dt=4.3 cm/s=0.043 m/s

==>r=0.043*t+c

where c is a constant

at t=0, r=c=0.30119

hence r=0.043*t+0.30119 m

then magnitude of emf induced =2*pi*0.28*(0.043*t+0.30119)*0.043

at t=0, emf induced=22.785 mV=23 mV(approx)

at t=1 seconds, emf induced=26.038 mV=26 mV(approx)

Related Questions

Navigate

• Browse (All)
• Browse I
• Subjects

---