Answer:
the magnitude of the induced current is 182.86 A.
Step-by-step explanation:
Given;
number of turns, N = 240 turns
cross sectional area of the loop, A = 0.2 m²
uniform magnetic field strength, B = 1.6 T
resistance of the loop, R = 21 ohms
time, Δt = 20.0 ms
The magnitude of the induced emf is calculated as;
![emf = (NA B)/(t) \\\\emf = (240 * 0.2 * 1.6)/(20 * 10^(-3)) \\\\emf = 3,840\ V](https://img.qammunity.org/2022/formulas/physics/college/covrntdrdwl2ybtjxx7am0rotnda77abrm.png)
The induced current in the loop is calculated as;
![I = (emf)/(R) \\\\I = (3840)/(21) \\\\I= 182.86 \ A](https://img.qammunity.org/2022/formulas/physics/college/kh7svv38e9vdm66r69vzmphkagxn9i7f75.png)
Therefore, the magnitude of the induced current is 182.86 A.