A flat loop of wire consisting of a single turn of crosssect

A flat loop of wire consisting of a single turn of cross-sectional area 7.00 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 2.90 T in 0.98 s. What is the resulting induced current if the loop has a resistance of 1.90 ?

Solution

the induced emf is equal to negative the change in flux over time where the flux is B*A
so:
e = -(Bf * A - Bi * A) / (Tf - Ti)

Bf = 2.9 T
Bi = 0.5 T
A = 7. cm^2 = 0.00070 m^2
Tf = .98 s
Ti = 0 s

e = -(2.9 * 0.00070 - 0.5 * 0.00070) / (.98)
= 0.00171 V

ohm\'s law states:
e = ir
therefore
i = e/r
= 0.00171 / 1.90
= 0.00090 A

which is 0.9 mA