The image contains a physics problem about electromagnetic induction. ### Problem Statement: The magnetic flux through the loop in the accompanying figure varies with time according to the equation: \[ \Phi_m = \Phi_0 e^{-\alpha t} \sin(\omega t), \] where: - $\Phi_0 = 4.5 \times 10^{-3} \, \text{T} \cdot \text{m}^2$, - $\alpha = 3 \, \text{s}^{-1}$, - $\omega = 120 \pi \, \text{rad/s}$. The task is to determine the direction and magnitude of the current through a 5.0 Ω resistor at the following times: - (a) $t = 0$, - (b) $t = 2.17 \times 10^{-2} \, \text{s}$, - (c) $t = 3.00 \, \text{s}$. ### Diagram Explanation: - The diagram shows a loop of wire with a resistor of 5.0 Ω connected across it. - A uniform magnetic field, denoted by $\mathbf{B}$, is represented by dots indicating the direction is perpendicular to the plane of the loop. ### Provided Hints: - (a) Current flows down with 339 mA at $t = 0$. - (b) Current flows up with 0.1734 mA at $t = 2.17 \times 10^{-2} \, \text{s}$. - (c) Current flows down with 0.042 mA at $t = 3.00 \, \text{s}$. ### Interface Options: - Dropdown menus to select the direction of current flow. - Text fields to input the magnitude of current in mA at each specified time. ### Additional Interface Features: - Buttons for "Question Help" and "Submit Question". - A visual computer interface is seen in the background, identified as a MacBook Air keyboard. This problem aids in understanding electromagnetic induction concepts, notably how a changing magnetic field induces current in a loop.

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Part b was wrong question 3 please show all work

The image contains a physics problem about electromagnetic induction.

### Problem Statement:
The magnetic flux through the loop in the accompanying figure varies with time according to the equation:

\[
\Phi_m = \Phi_0 e^{-\alpha t} \sin(\omega t),
\]

where:
- $\Phi_0 = 4.5 \times 10^{-3} \, \text{T} \cdot \text{m}^2$,
- $\alpha = 3 \, \text{s}^{-1}$,
- $\omega = 120 \pi \, \text{rad/s}$.

The task is to determine the direction and magnitude of the current through a 5.0 Ω resistor at the following times:
- (a) $t = 0$,
- (b) $t = 2.17 \times 10^{-2} \, \text{s}$,
- (c) $t = 3.00 \, \text{s}$.

### Diagram Explanation:
- The diagram shows a loop of wire with a resistor of 5.0 Ω connected across it.
- A uniform magnetic field, denoted by $\mathbf{B}$, is represented by dots indicating the direction is perpendicular to the plane of the loop.

### Provided Hints:
- **(a)** Current flows down with 339 mA at $t = 0$.
- **(b)** Current flows up with 0.1734 mA at $t = 2.17 \times 10^{-2} \, \text{s}$.
- **(c)** Current flows down with 0.042 mA at $t = 3.00 \, \text{s}$.

### Interface Options:
- Dropdown menus to select the direction of current flow.
- Text fields to input the magnitude of current in mA at each specified time.

### Additional Interface Features:
- Buttons for "Question Help" and "Submit Question".
- A visual computer interface is seen in the background, identified as a MacBook Air keyboard.

This problem aids in understanding electromagnetic induction concepts, notably how a changing magnetic field induces current in a loop.

Expert Solution

Step 1

Given,

magnetic flux

$ϕ = ϕ_{0} e^{- α t} \sin (ω t)$

Step 2

The induced emf will be

$ε = - \frac{d ϕ}{d t} ε = - \frac{d}{d t} (ϕ_{0} e^{- α t} \sin (ω t)) ε = - ϕ_{0} (e^{- α t} \cos (ω t) ω + \sin (ω t) * e^{- α t} * (- α)) ε = ϕ_{0} e^{- α t} (α \sin ω t - ω \cos ω t)$

then induced current is

$I = \frac{ε}{R} I = \frac{ϕ_{0} e^{- α t} (α \sin ω t - ω \cos ω t)}{R} I = \frac{4.5 \times 10^{- 3}}{5} e^{- 3 t} (3 \sin (120 π t) - 120 π \cos (120 π t)) I = 0.9 \times 10^{- 3} * e^{- 3 t} (3 \sin (120 π t) - 120 π \cos (120 π t))$

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