https://file.aiquickdraw.com/mathbot/file/173025785475628eyjdes.jpg

Let’s work through the problem in **Part C** by calculating the angular acceleration \( \alpha \) of the coil just after the current is started. We are given the following information:

- Mass of the coil \( m = 240 \, \text{g} = 0.24 \, \text{kg} \)
- Dimensions of the coil: \( 0.500 \, \text{m} \times 1.00 \, \text{m} \)
- Magnetic field \( B = 3.50 \, \text{T} \)
- Current \( I = 2.10 \, \text{A} \)
- Axis of rotation: \( A_2 \), which is perpendicular to the 1.00 m side of the coil.

### Step 1: Calculate the Torque on the Coil
The magnetic torque \( \tau \) on a current-carrying coil in a magnetic field is given by:
\[
\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta)
\]
where:
- \( n = 1 \) (since it’s a single loop),
- \( I = 2.10 \, \text{A} \),
- \( A = \text{area of the coil} = 0.500 \, \text{m} \times 1.00 \, \text{m} = 0.5 \, \text{m}^2 \),
- \( B = 3.50 \, \text{T} \),
- \( \theta = 90^\circ \) (since the coil is parallel to the magnetic field, so \( \sin(90^\circ) = 1 \)).

Plugging in these values:
\[
\tau = 1 \cdot 2.10 \, \text{A} \cdot 0.5 \, \text{m}^2 \cdot 3.50 \, \text{T} \cdot 1 = 3.675 \, \text{N} \cdot \text{m}
\]

### Step 2: Calculate the Moment of Inertia of the Coil
Since the coil rotates about the axis \( A_2 \), we treat it as a thin rectangular loop rotating about one side. The moment of inertia \( I_{\text{coil}} \) of a rectangle about an axis along one of its sides (width \( w = 0.500 \, \text{m} \) and length \( L = 1.00 \, \text{m} \)) is:
\[
I_{\text{coil}} = \frac{1}{3} m L^2
\]
Substituting the values:
\[
I_{\text{coil}} = \frac{1}{3} \cdot 0.24 \, \text{kg} \cdot (1.00 \, \text{m})^2 = 0.08 \, \text{kg} \cdot \text{m}^2
\]

### Step 3: Calculate the Angular Acceleration
Using the relationship \( \tau = I_{\text{coil}} \cdot \alpha \), we can solve for the angular acceleration \( \alpha \):
\[
\alpha = \frac{\tau}{I_{\text{coil}}} = \frac{3.675 \, \text{N} \cdot \text{m}}{0.08 \, \text{kg} \cdot \text{m}^2} = 45.94 \, \text{rad/s}^2
\]

### Answer
The initial angular acceleration \( \alpha \) of the coil just after the current is started is approximately:
\[
\alpha \approx 45.94 \, \text{rad/s}^2
\]

If this answer does not match, please double-check that all input values are correct.

Would you like further details on any of these steps?

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**Additional Practice Questions:**
1. How does the torque change if the coil were placed at an angle other than \(90^\circ\) to the magnetic field?
2. What would be the moment of inertia if the coil had different dimensions?
3. How would changing the current affect the initial angular acceleration?
4. What happens to the angular acceleration if the magnetic field strength is doubled?
5. How would the angular acceleration change if the coil mass were doubled?

**Tip:** Remember that torque depends on the angle between the current loop and the magnetic field.

Find the initial angular acceleration of a uniform rectangular coil with given mass, dimensions, magnetic field, and current.

Learn how to find the initial angular acceleration of a rectangular coil in a magnetic field, given the coil's mass, dimensions, current, and magnetic field strength. This problem covers key physics concepts including torque on a coil, moment of inertia, and angular acceleration, suitable for college-level physics students.

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