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n a side. The solenoid is 0.20 cm long and wound with 200 turns of wire. If the current in the solenoid is 3.00 A, what is the flux through the square loop? If the current in the solenoid

n a side. The solenoid is 0.20 cm long and wound with 200 turns of wire. If the current in the solenoid is 3.00 A, what is the flux through the square loop? If the current in the solenoid

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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A single square loop of wire is placed in the center of a large solenoid as shown. The solenoid has a radius 5.00 cm and the square loop is 3.0 cm on a side. The solenoid is 0.20 cm long and wound with 200 turns of wire. If the current in the solenoid is 3.00 A, what is the flux through the square loop? If the current in the solenoid drops to zero in 0.500 seconds, what is the magnitude of the average induced emf in the square loop? 

### Physics Education: Induced EMF in a Solenoid **Problem Statement:** A single square loop of wire is placed in the center of a large solenoid as shown. The solenoid has a radius of 5.00 cm and is 20 cm long with 200 turns of wire. The side of the square loop is 3.00 cm. The current in the solenoid is 3.00 A. 1. What is the flux through the square loop? 2. If the current in the solenoid drops to zero in 0.500 s, what is the magnitude of the average induced emf in the square loop? **Diagram Explanation:** The diagram consists of a large outer circle representing the cross-section of the solenoid and a smaller square inside the circle representing the square loop. The arrangement indicates how the square loop is positioned centrally within the solenoid. ### Analysis: 1. **Magnetic Flux Calculation:** - To find the magnetic flux through the square loop, we can use the formula for magnetic flux \( \Phi = B \cdot A \), where \( B \) is the magnetic field inside the solenoid and \( A \) is the area of the square loop. - The magnetic field inside a solenoid is given by \( B = \mu_0 \cdot (n \cdot I) \) with \( n \) being the number of turns per unit length and \( I \) the current. 2. **Average Induced EMF:** - According to Faraday's Law of Electromagnetic Induction, the induced emf \( \varepsilon \) is given by \( \varepsilon = - \Delta \Phi / \Delta t \), where \( \Delta \Phi \) is the change in magnetic flux and \( \Delta t \) is the time interval for the change (0.500 s). The above outlines the basic approach needed to solve for the magnetic flux and the average induced emf in the given scenario. These calculations require an understanding of electromagnetism and the properties of solenoids and loops.
Transcribed Image Text:### Physics Education: Induced EMF in a Solenoid **Problem Statement:** A single square loop of wire is placed in the center of a large solenoid as shown. The solenoid has a radius of 5.00 cm and is 20 cm long with 200 turns of wire. The side of the square loop is 3.00 cm. The current in the solenoid is 3.00 A. 1. What is the flux through the square loop? 2. If the current in the solenoid drops to zero in 0.500 s, what is the magnitude of the average induced emf in the square loop? **Diagram Explanation:** The diagram consists of a large outer circle representing the cross-section of the solenoid and a smaller square inside the circle representing the square loop. The arrangement indicates how the square loop is positioned centrally within the solenoid. ### Analysis: 1. **Magnetic Flux Calculation:** - To find the magnetic flux through the square loop, we can use the formula for magnetic flux \( \Phi = B \cdot A \), where \( B \) is the magnetic field inside the solenoid and \( A \) is the area of the square loop. - The magnetic field inside a solenoid is given by \( B = \mu_0 \cdot (n \cdot I) \) with \( n \) being the number of turns per unit length and \( I \) the current. 2. **Average Induced EMF:** - According to Faraday's Law of Electromagnetic Induction, the induced emf \( \varepsilon \) is given by \( \varepsilon = - \Delta \Phi / \Delta t \), where \( \Delta \Phi \) is the change in magnetic flux and \( \Delta t \) is the time interval for the change (0.500 s). The above outlines the basic approach needed to solve for the magnetic flux and the average induced emf in the given scenario. These calculations require an understanding of electromagnetism and the properties of solenoids and loops.
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