A single loop of copper wire, lying flat in a plane, has an area of 7.20 cm? and a resistance of 2.10 Q. A uniform magnetic field points perpendicular to the plane of the loop. The field initially has a magnitude of 0.500 T, and the magnitude increases linearly to 2.00 T in a time of 1.06 s. What is the induced current (in mA) in the loop of wire over this time? mA
A single loop of copper wire, lying flat in a plane, has an area of 7.20 cm? and a resistance of 2.10 Q. A uniform magnetic field points perpendicular to the plane of the loop. The field initially has a magnitude of 0.500 T, and the magnitude increases linearly to 2.00 T in a time of 1.06 s. What is the induced current (in mA) in the loop of wire over this time? mA
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![**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 Ω
A 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:**
\[ \Delta \](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3d3b999f-40bc-4397-b918-1b796d12487f%2F601a0a71-ac1e-4cb3-be79-f5f8c69abb01%2F7emv1l_processed.png&w=3840&q=75)
Transcribed Image Text:**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 Ω
A 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 \
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