Math Problem Statement

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

Solution

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

  • Mass of the coil m=240g=0.24kgm = 240 \, \text{g} = 0.24 \, \text{kg}
  • Dimensions of the coil: 0.500m×1.00m0.500 \, \text{m} \times 1.00 \, \text{m}
  • Magnetic field B=3.50TB = 3.50 \, \text{T}
  • Current I=2.10AI = 2.10 \, \text{A}
  • Axis of rotation: A2A_2, which is perpendicular to the 1.00 m side of the coil.

Step 1: Calculate the Torque on the Coil

The magnetic torque τ\tau on a current-carrying coil in a magnetic field is given by: τ=nIABsin(θ)\tau = n \cdot I \cdot A \cdot B \cdot \sin(\theta) where:

  • n=1n = 1 (since it’s a single loop),
  • I=2.10AI = 2.10 \, \text{A},
  • A=area of the coil=0.500m×1.00m=0.5m2A = \text{area of the coil} = 0.500 \, \text{m} \times 1.00 \, \text{m} = 0.5 \, \text{m}^2,
  • B=3.50TB = 3.50 \, \text{T},
  • θ=90\theta = 90^\circ (since the coil is parallel to the magnetic field, so sin(90)=1\sin(90^\circ) = 1).

Plugging in these values: τ=12.10A0.5m23.50T1=3.675Nm\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 A2A_2, we treat it as a thin rectangular loop rotating about one side. The moment of inertia IcoilI_{\text{coil}} of a rectangle about an axis along one of its sides (width w=0.500mw = 0.500 \, \text{m} and length L=1.00mL = 1.00 \, \text{m}) is: Icoil=13mL2I_{\text{coil}} = \frac{1}{3} m L^2 Substituting the values: Icoil=130.24kg(1.00m)2=0.08kgm2I_{\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 τ=Icoilα\tau = I_{\text{coil}} \cdot \alpha, we can solve for the angular acceleration α\alpha: α=τIcoil=3.675Nm0.08kgm2=45.94rad/s2\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 α\alpha of the coil just after the current is started is approximately: α45.94rad/s2\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?


Additional Practice Questions:

  1. How does the torque change if the coil were placed at an angle other than 9090^\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.

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Math Problem Analysis

Mathematical Concepts

Electromagnetism
Rotational Dynamics
Angular Acceleration

Formulas

Torque τ = n * I * A * B * sin(θ)
Moment of Inertia for a rectangular loop I_coil = (1/3) * m * L^2
Angular Acceleration α = τ / I_coil

Theorems

Newton's Second Law for Rotation

Suitable Grade Level

College-level Physics