Answer :
Complete 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 1.60 T in 0.99 s. What is the resulting induced current if the loop has a resistance of [tex]1.20 \ \Omega[/tex]
Answer:
The current is [tex]I = 0.0007 41 \ A[/tex]
Explanation:
From the question we are told that
The area is [tex]A = 8.00 \ cm^2 = 8.0 *10^{-4} \ m^2[/tex]
The initial magnetic field at [tex]t_o = 0 \ seconds[/tex] is [tex]B_i = 0.500 \ T[/tex]
The magnetic field at [tex]t_1 = 0.99 \ seconds[/tex] is [tex]B_f = 1.60 \ T[/tex]
The resistance is [tex]R = 1.20 \ \Omega[/tex]
Generally the induced emf is mathematically represented as
[tex]\epsilon = A * \frac{B_f - B_i }{ t_f - t_o }[/tex]
=> [tex]\epsilon = 8.0 *10^{-4} * \frac{1.60 - 0.500 }{ 0.99- 0 }[/tex]
=> [tex]\epsilon = 0.000889 \ V[/tex]
Generally the current induced is mathematically represented as
[tex]I = \frac{\epsilon}{R }[/tex]
=> [tex]I = \frac{0.000889}{ 1.20 }[/tex]
=> [tex]I = 0.0007 41 \ A[/tex]