To solve the problem, we need to find the acceleration of the wire carrying a current in a magnetic field. Here are the steps to arrive at the solution:
### Step 1: Identify the Given Values
- Mass of the wire, \( m = 100 \, \text{g} = 0.1 \, \text{kg} \)
- Current in the wire, \( I = 2 \, \text{A} \)
- Magnetic field, \( \mathbf{B} = -0.02 \hat{k} \, \text{T} \)
### Step 2: Determine the Effective Length of the Wire
The wire is described by the equation \( y = x^2 \) from \( x = -2 \) to \( x = 2 \). The effective length of the wire can be calculated as follows:
- The length of the wire can be calculated using the limits of integration:
\[
L = \int_{-2}^{2} \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]
The derivative \( \frac{dy}{dx} = 2x \). Therefore,
\[
L = \int_{-2}^{2} \sqrt{1 + (2x)^2} \, dx = \int_{-2}^{2} \sqrt{1 + 4x^2} \, dx
\]
However, for the sake of this problem, we can consider the effective length of the wire to be \( L = 4 \, \text{m} \) (the horizontal span from -2 to 2).
### Step 3: Calculate the Force on the Wire
The force \( \mathbf{F} \) on a current-carrying wire in a magnetic field is given by the formula:
\[
\mathbf{F} = I \mathbf{L} \times \mathbf{B}
\]
Where:
- \( \mathbf{L} = 4 \hat{i} \, \text{m} \) (since the wire is along the x-direction)
- \( \mathbf{B} = -0.02 \hat{k} \, \text{T} \)
Now, we calculate the cross product:
\[
\mathbf{F} = 2 \cdot (4 \hat{i}) \times (-0.02 \hat{k})
\]
Using the right-hand rule, we find:
\[
\hat{i} \times \hat{k} = \hat{j} \quad \Rightarrow \quad \mathbf{F} = 8 \cdot (-0.02) \hat{j} = -0.16 \hat{j} \, \text{N}
\]
### Step 4: Calculate the Acceleration
Using Newton's second law, \( F = ma \):
\[
a = \frac{F}{m}
\]
Substituting the values we have:
\[
a = \frac{-0.16 \hat{j}}{0.1} = -1.6 \hat{j} \, \text{m/s}^2
\]
### Final Answer
The acceleration of the wire is:
\[
\mathbf{a} = -1.6 \hat{j} \, \text{m/s}^2
\]
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