A flat loop of wire consisting of a single turn of cross-sectional area 8 cm2 is...
A flat loop of wire consisting of a single turn of cross-sectional area 8 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.5 T to 2.5 T in a time of 1.19 s. The total resistance is 0.719 Ω. What is the resulting induced current? Answer in units of mA.
Homework Answers
Given Data
No. of turns N = 1
change in magnetic field dB = B2 - B1
= 2.50 T - 0.500 T
= 2.00 T
Corresponding time dt = 1.19s
Area of loop A = 8.00 cm2 = 8.00 * 10-4 m2
Solution :-
Induced EMF E = - dφ/dt = - N * A * dB/dt
E = - N * A * dB/dt
E = - 1 * 8.00 * 10-4 * 2.00 / 1.19
= - 1.344 * 10-3 V
= - 1.344 mV
- ve sign can be discarded as it indicates direction only.
Current I = E / R
= 1.344 mV / 0.719 Ω
= 1.87 mA --> Answer
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