Answer:
(a) 1.58 V
(b) 0.0126 Wb
(c) 0.0493 V
Solution:
As per the question:
No. of turns in the coil, N = 400 turns
Self Inductance of the coil, L = 7.50 mH = [tex]7.50\times 10^{- 3}\ H[/tex]
Current in the coil, i = [tex]1680cos[\frac{\pi t}{0.0250}][/tex] A
where
[tex]i_{max} = 1680\ mA = 1.680\ A[/tex]
Now,
(a) To calculate the maximum emf:
We know that maximum emf induced in the coil is given by:
[tex]e = \frac{Ldi}{dt}[/tex]
[tex]e = L\frac{d}{dt}(1680)cos[\frac{\pi t}{0.0250}][/tex]
[tex]e = - 7.50\times 10^{- 3}\times \frac{\pi}{0.0250}\times \frac{d}{dt}(1680)sin[\frac{\pi t}{0.0250}][/tex]
For maximum emf, [tex]sin\theta[/tex] should be maximum, i.e., 1
Now, the magnitude of the maximum emf is given by:
[tex]|e| = 7.50\times 10^{- 3}\times 1680\times 10^{- 3}\times \frac{\pi}{0.0250} = 1.58\ V[/tex]
(b) To calculate the maximum average flux,we know that:
[tex]\phi_{m, avg} = L\times i_{max} = 7.50\times 10^{- 3}\times 1.680 = 0.0126\ Wb[/tex]
(c) To calculate the magnitude of the induced emf at t = 0.0180 s:
[tex]e = e_{o}sin{\pi t}{0.0250}[/tex]
[tex]e = 7.50\times 10^{- 3}\times sin{\pi \times 0.0180}{0.0250} = 2.96\times 10^{- 4} =0.0493\ V[/tex]
