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A flat loop of wire consisting of a single turn of cross-sectional area 8.80 cm^2 is perpendicular to a magnetic field that increases unifor
Question
A flat loop of wire consisting of a single turn of cross-sectional area 8.80 cm^2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.80 T in 1.01 s.
a. What is the resulting induced current if the loop has a resistance of 1.00Ω?
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Physics
3 years
2021-08-09T14:41:31+00:00
2021-08-09T14:41:31+00:00 1 Answers
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Answers ( )
Answer:
0.00122 A
Explanation:
Parameters given:
Area, A = 8.8 cm² = 0
00088 m²
Initial Magnetic field, Bin = 0.5 T
Final magnetic field, Bfin = 2.8
Time taken, t = 1.01 s
Resistance, R = 1 ohms
EMF induced in a loop of wire of cross sectional area, A due to a changing magnetic field is:
EMF = [-(Bfin – Bin) * N * A] / t
Where N = number of loops = 1
EMF = [-(2.8 – 0.5) * 1 * 0.00088] / 1.01
EMF = -0.00122 V
We know that
V = I * R
Hence, current, I, will be:
I = V/R
I = 0.00122/1 = 0.00122 A