To determine the induced current in the loop of wire over time, we can apply Faraday’s law of electromagnetic induction. This law states that the induced electromotive force (emf) in a loop of wire is equal to the rate of change of magnetic flux through the loop.

First, let’s calculate the initial magnetic flux through the loop when the magnetic field magnitude is 0.500 T. The magnetic flux (Φ) is defined as the product of the magnetic field magnitude (B) and the area (A) of the loop:

Φinitial = Binitial * A

Given that the area of the loop is 7.20 cm^2, we need to convert it to square meters to have consistent units. 1 cm^2 is equal to 1 * 10^-4 m^2, so the area in square meters (A) is:

A = 7.20 cm^2 * (1 * 10^-4 m^2 / 1 cm^2) = 7.20 * 10^-4 m^2

Plugging in the values, we have:

Φinitial = 0.500 T * 7.20 * 10^-4 m^2 = 3.60 * 10^-4 Wb

Next, let’s calculate the final magnetic flux through the loop when the magnetic field magnitude is 2.00 T. Using the same formula as above:

Φfinal = Bfinal * A

Plugging in the values, we have:

Φfinal = 2.00 T * 7.20 * 10^-4 m^2 = 1.44 * 10^-3 Wb

Now, we can calculate the rate of change of magnetic flux (∆Φ/∆t). Since the magnetic field magnitude increases linearly from 0.500 T to 2.00 T in a time of 1.06 s, we can divide the change in magnetic flux (∆Φ) by the change in time (∆t) to obtain the rate of change:

∆Φ/∆t = (Φfinal – Φinitial) / ∆t

Plugging in the values, we have:

∆Φ/∆t = (1.44 * 10^-3 Wb – 3.60 * 10^-4 Wb) / 1.06 s

Simplifying the expression gives us the rate of change of magnetic flux:

∆Φ/∆t = 1.08 * 10^-3 Wb / 1.06 s = 1.02 * 10^-3 Wb/s

According to Faraday’s law of electromagnetic induction, the induced emf (ε) is equal to the negative of the rate of change of magnetic flux (∆Φ/∆t). Therefore:

ε = -∆Φ/∆t = -1.02 * 10^-3 Wb/s

The induced emf is related to the induced current (I) and the resistance (R) of the loop according to Ohm’s law:

ε = I * R

Solving for I, we have:

I = ε / R

Plugging in the values, we have:

I = (-1.02 * 10^-3 Wb/s) / (2.10 Ω)

Now, we need to convert the current to milliamperes (mA). 1 A is equal to 1000 mA, so the induced current in milliamperes is:

I (in mA) = I (in A) * 1000

Plugging in the values, we have:

I (in mA) = (-1.02 * 10^-3 A) / (2.10 Ω) * 1000 = -0.486 mA

Therefore, the induced current in the loop of wire over the given time is approximately -0.486 mA.

Please note that the negative sign indicates that the induced current flows in the opposite direction to the conventional current flow.