**Induced Current Calculation in a Copper Loop****Problem Statement:**A single loop of copper wire, lying flat in a plane, has the following characteristics:- **Area:** 7.20 cm²- **Resistance:** 2.10 ΩA uniform magnetic field is oriented perpendicular to the plane of the loop. The magnetic field's initial magnitude is 0.500 T and increases linearly to 2.00 T within a time span of 1.06 s. **Question:** What is the induced current in milliamperes (mA) in the loop of wire over this time?**Calculation Detail:**To calculate the induced current (I) in the loop, we can use Faraday's Law of Electromagnetic Induction and Ohm's Law.**1. Faraday's Law:** \( \mathcal{E} = -\frac{d\Phi_B}{dt} \) Here, \( \mathcal{E} \) is the induced EMF (Electromotive Force), and \( \Phi_B \) is the magnetic flux, given by \( \Phi_B = B \cdot A \cdot \cos(\theta) \). Since the magnetic field is perpendicular to the loop, \(\cos(\theta) = \cos(0) = 1\), so \( \Phi_B = B \cdot A \).**Change in Magnetic Flux \( (\Delta \Phi_B) \):** \( \Delta \Phi_B = A \cdot \Delta B = A \cdot (B_{\text{final}} - B_{\text{initial}}) \)Here, \[ A = 7.20 \, \text{cm}^2 = 7.20 \times 10^{-4} \, \text{m}^2 \] \[ B_{\text{final}} = 2.00 \, \text{T} \] \[ B_{\text{initial}} = 0.500 \, \text{T} \] Therefore, \[ \Delta B = B_{\text{final}} - B_{\text{initial}} = 2.00 \, \text{T} - 0.500 \, \text{T} = 1.50 \, \text{T} \]**Magnetic Flux Change:** \[ \Delta \

Area of copper wire A = 7.20 cm2 = 7.20 × 10- 4 m2 Resistance of copper wire R = 2.10 Ω Initial…

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