Question

A flat loop of wire consisting of a single turn of cross-sectional area 8.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.50 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 2.00 W?

Answer 1

Answer:-

E=(Bf-Bi)*A/t=(2.5-0.5)*8*10-4

/1=1.6*10-3 V

I=E/R=1.23 mA

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