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 2.20 t in 1.08 s. what is the resulting induced current if the loop has a resistance of 2.60 ?

Answer 1

The area enclosed by the wire is
$A=8.00 cm^2 = 8.00 \cdot 10^{-4} m^2$

By using Faraday-Neumann-Lenz law, we can find the emf induced in the circuit, whose magnitude is equal to the variation of magnetic flux on the wire:
$\epsilon = \frac{\Delta \Phi}{\Delta t}= \frac{\Delta B A}{\Delta t}= \frac{(2.2 T-0.5 T)(8 \cdot 10^{-4} m^2)}{1.08 s}=1.26 \cdot 10^{-3} V$

And then, by using Ohm's law, we can find the induced current:
$I= \frac{\epsilon}{R}= \frac{1.26 \cdot 10^{-3} V}{2.60 \Omega}=4.86 \cdot 10^{-4}A$