A small, 2.00-mm-diameter circular loop with 1.30×10-2 is at the center of a large 110-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from + 1.0 A to - 1.0 A in 9.00×10-² s.

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ISBN:9781305952300
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Chapter1: Units, Trigonometry. And Vectors
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### Problem Description

A small, 2.00-mm-diameter circular loop with a resistance of \(1.30 \times 10^{-2} \, \Omega\) is located at the center of a large 110-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from \(+1.0 \, \text{A}\) to \(-1.0 \, \text{A}\) in \(9.00 \times 10^{-2} \, \text{s}\).

### Part A

**Question:** What is the induced current in the inner loop?  
**Instruction:** Express your answer with the appropriate units.

**Response Format:**
The induced current \(I\) should be calculated and entered with the correct units.

**Additional Features:**
- There is an option to view hints to assist with solving the problem.
- Input fields are provided for entering the value and units of the induced current.
Transcribed Image Text:### Problem Description A small, 2.00-mm-diameter circular loop with a resistance of \(1.30 \times 10^{-2} \, \Omega\) is located at the center of a large 110-mm-diameter circular loop. Both loops lie in the same plane. The current in the outer loop changes from \(+1.0 \, \text{A}\) to \(-1.0 \, \text{A}\) in \(9.00 \times 10^{-2} \, \text{s}\). ### Part A **Question:** What is the induced current in the inner loop? **Instruction:** Express your answer with the appropriate units. **Response Format:** The induced current \(I\) should be calculated and entered with the correct units. **Additional Features:** - There is an option to view hints to assist with solving the problem. - Input fields are provided for entering the value and units of the induced current.
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