Answer:
a) Magnitude of maximum emf induced = 0.0714 V = 71.4 mv
b) Maximum current through the bulb = 0.00793 A = 7.93 mA
Explanation:
a) The induced emf from Faraday's law of electromagnetic induction is related to angular velocity through
E = NABw sin wt
The maximum emf occurs when (sin wt) = 1
Maximum Emf = NABw
N = 1
A = 4 cm² = 0.0004 m²
B = 6 T
w = (284/60) × 2π = 29.75 rad/s
E(max) = 1×0.0004×6×29.75 = 0.0714 V = 71.4 mV
Note that: since we're after only the magnitude of the induced emf, the minus sign that indicates that the induced emf is 8n the direction opposite to the change in magnetic flux, is ignored for this question.
b) Maximum current through the bulb
E(max) = I(max) × R
R = 9 ohms
E(max) = 0.0714 V
I(max) = ?
0.0714 = I(max) × 9
I(max) = (0.0714/9) = 0.00793 A = 7.93 mA
Hope this Helps!!
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