Answer:
the magnitude of the induced current is 182.86 A.
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;
[tex]emf = \frac{NA B}{t} \\\\emf = \frac{240 \times 0.2 \times 1.6}{20 \times 10^{-3}} \\\\emf = 3,840\ V[/tex]
The induced current in the loop is calculated as;
[tex]I = \frac{emf}{R} \\\\I = \frac{3840}{21} \\\\I= 182.86 \ A[/tex]
Therefore, the magnitude of the induced current is 182.86 A.