Final answer:
The (a) maximum emf induced in the coil is -12.60 V. (b) The maximum average flux through each turn of the coil is -31.5 mV. (c) At t=0.0180s, the magnitude of the induced emf is 12.76 V.
Step-by-step explanation:
(a) To find the maximum emf induced in the coil, we need to find the maximum value of the current. The maximum value of the current can be found by substituting the maximum value of t, which is 0.0250s, into the equation for i:
i = 1680 mA^2 cos [πt/(0.0250s)]
Substituting t = 0.0250s, we get:
i = 1680 mA^2 cos [π(0.0250s)/(0.0250s)]
i = 1680 mA^2 cos [π]
Since cos[π] = -1, the maximum value of the current is:
i = 1680 mA^2 * (-1)
i = -1680 mA^2
The maximum emf induced in the coil is the product of the maximum value of the current and the self-inductance of the coil:
Emf = (Max current) * (Self-inductance)
Emf = (-1680 mA^2) * (7.50 mH)
Emf = -12.60 V
(b) The maximum average flux through each turn of the coil can be found by dividing the maximum emf by the number of turns:
Max average flux per turn = Max emf / Number of turns
Max average flux per turn = -12.60 V / 400 turns
Max average flux per turn = -31.5 mV
(c) To find the magnitude of the induced emf at t = 0.0180s, we can substitute t = 0.0180s into the equation for i:
i = 1680 mA^2 cos [πt/(0.0250s)]
Substituting t = 0.0180s, we get:
i = 1680 mA^2 cos [π(0.0180s)/(0.0250s)]
i = 1680 mA^2 cos [π(0.72)]
Since cos[π(0.72)] = -0.953, the magnitude of the induced emf is:
Emf = (Magnitude of current) * (Self-inductance)
Emf = (1680 mA^2 * 0.953) * (7.50 mH)
Emf = 12.76 V