Answer:
The magnitude of the induced current is 4.73 x 10⁻³ A.
Explanation:
Given;
number of turns, N = 1
cross sectional area of the loop, A = 8.8 cm² = 8.8 x 10⁻⁴ m²
change in magnetic field strength, ΔB = 1.8 T - 0.5 T = 1.3 T
change in time, Δt = 1.10 s
resistance of the loop, R = 2.2 ohm
The magnitude of the induced emf is calculated as;
[tex]emf = \frac{NA \Delta B}{\Delta t} \\\\emf = \frac{1 \times 8.8\times 10^{-4} \times 1.3}{1.10} \\\\emf = 1.04 \times 10^{-3} \ V[/tex]
The induced current in the loop is calculated as;
[tex]I = \frac{emf}{R} \\\\I = \frac{1.04 \times 10^{-3}}{2.2} \\\\I= 4.73 \times 10^{-4} \ A[/tex]
Therefore, the magnitude of the induced current is 4.73 x 10⁻³ A
