Let's break this down step by step, solving each part of the problem.

Given:

• Current in the wire: $I(t) = (8 - t^2) \, \text{A}$
• Dimensions of the rectangular loop: $4 \, \text{cm} \times 6 \, \text{cm}$
• Distance from the wire to the loop: $2 \, \text{cm}$
• Resistance of the loop: $R = 10 \, \Omega$
• Find the quantities at $t = 2 \, \text{s}$

1. Direction of the Magnetic Field in the Loop

The magnetic field around a current-carrying wire is given by the right-hand rule. Point your thumb in the direction of the current, and your fingers will curl in the direction of the magnetic field. Since the current in the straight wire is directed upwards (as shown in the diagram), the magnetic field around the wire will circulate in a clockwise direction around the wire.

At the position of the rectangular loop to the right of the wire:

• On the left side of the loop (closer to the wire), the magnetic field will point into the page.
• On the right side of the loop (farther from the wire), the magnetic field will point out of the page.

2. Magnetic Flux Through the Loop Due to the Current

The magnetic field at a distance $r$ from a long straight wire carrying current $I$ is: $B(r) = \frac{\mu_0 I}{2 \pi r}$ where $\mu_0 = 4\pi \times 10^{-7} \, \text{T·m/A}$ is the permeability of free space.

The magnetic flux $\Phi$ through the loop is the integral of the magnetic field over the area of the loop: $\Phi = \int_{r_1}^{r_2} B(r) \, \mathrm{d}A = \int_{r_1}^{r_2} \frac{\mu_0 I}{2\pi r} \cdot h \, \mathrm{d}r$ Here:

• $h = 6 \, \text{cm} = 0.06 \, \text{m}$ is the height of the loop.
• $r_1 = 2 \, \text{cm} = 0.02 \, \text{m}$ is the distance from the wire to the left edge of the loop.
• $r_2 = 4 \, \text{cm} + 2 \, \text{cm} = 0.06 \, \text{m}$ is the distance from the wire to the right edge of the loop.

Substituting the values, we can calculate the magnetic flux: $\Phi = \frac{\mu_0 I h}{2\pi} \int_{r_1}^{r_2} \frac{1}{r} \, \mathrm{d}r$ $\Phi = \frac{\mu_0 I h}{2\pi} \ln\left(\frac{r_2}{r_1}\right)$

3. Induced EMF in the Loop

Faraday's law of induction states that the induced emf $\mathcal{E}$ in a loop is the negative rate of change of magnetic flux through the loop: $\mathcal{E} = -\frac{\mathrm{d}\Phi}{\mathrm{d}t}$

Since the current $I(t)$ is changing with time, the flux through the loop also changes. The rate of change of the current is: $\frac{\mathrm{d}I}{\mathrm{d}t} = -2t$ At $t = 2 \, \text{s}$: $\frac{\mathrm{d}I}{\mathrm{d}t} = -4 \, \text{A/s}$ Using this, the induced emf can be found by differentiating the expression for $\Phi$ with respect to time.

4. Magnitude and Direction of the Induced Current at $t = 2 \, \text{s}$

The induced current is given by Ohm’s law: $I_{\text{induced}} = \frac{\mathcal{E}}{R}$ Using the emf calculated in the previous step, we can find the magnitude of the induced current.

Direction of the induced current: Lenz’s law tells us that the induced current will oppose the change in magnetic flux. Since the current in the straight wire is decreasing at $t = 2 \, \text{s}$, the magnetic flux through the loop is decreasing, and the induced current will flow in a direction that opposes this reduction. Therefore, the induced current will circulate to create a magnetic field in the same direction as the original field, which can be determined based on the right-hand rule.

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