Ch 30: Electromagnetic Induction

Knight Calc - Physics for Scientists and Engineers 5th Edition

Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796당신이 사용하는 게 아니라요?교과서 변경

Knight Calc 5th Edition

Ch 30: Electromagnetic Induction

문제 21

30장, 문제 21

A 12-cm-diameter, 1.0-m-long solenoid is wound with 2000 turns of superconducting wire. When the magnet is turned on, the current increases from 0 to I_max in 2.5 s. At t = 1.0 s, the induced electric field midway between the axis and the windings is 7.5×10⁻³ V/m. What is the solenoid's steady magnetic field strength?

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Determine the radius of the solenoid. The diameter is given as 12 cm, so the radius is half of that: \( r = \frac{12}{2} = 6 \; \text{cm} = 0.06 \; \text{m} \).

Use Faraday's law of electromagnetic induction to relate the induced electric field \( E \) to the rate of change of the magnetic flux. The formula is \( E = \frac{1}{2} r \frac{dB}{dt} \), where \( r \) is the radial distance from the axis, and \( \frac{dB}{dt} \) is the rate of change of the magnetic field.

Rearrange the formula to solve for \( \frac{dB}{dt} \): \( \frac{dB}{dt} = \frac{2E}{r} \). Substitute \( E = 7.5 \times 10^{-3} \; \text{V/m} \) and \( r = 0.06 \; \text{m} \) into the equation.

The steady magnetic field strength \( B \) is related to the maximum current \( I_{\text{max}} \) in the solenoid. Use the formula for the magnetic field inside a solenoid: \( B = \mu_0 n I \), where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \; \text{T·m/A} \)), \( n \) is the number of turns per unit length (\( n = \frac{N}{L} \)), and \( I \) is the current.

Combine the results: Use the value of \( \frac{dB}{dt} \) to find \( I_{\text{max}} \) and then calculate \( B \) using the solenoid formula. Note that \( N = 2000 \), \( L = 1.0 \; \text{m} \), and \( n = \frac{2000}{1.0} = 2000 \; \text{turns/m} \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Magnetic Field in a Solenoid

The magnetic field inside a long solenoid can be calculated using the formula B = μ₀(nI), where B is the magnetic field strength, μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. For a solenoid, the field is uniform and directed along the axis of the coil, making it essential for understanding how the solenoid generates a magnetic field when current flows through it.

Induced Electric Field

An induced electric field arises in a region when there is a change in magnetic flux over time, as described by Faraday's Law of Electromagnetic Induction. The induced electric field (E) can be related to the rate of change of the magnetic field (dB/dt) and the distance from the axis of the solenoid. This concept is crucial for analyzing the behavior of the electric field within the solenoid as the current changes.

Time-Varying Current

In this scenario, the current in the solenoid increases from 0 to I_max over a specified time period, which affects both the magnetic field and the induced electric field. The rate of change of current (dI/dt) is significant because it influences the strength of the induced electric field and the overall magnetic field generated by the solenoid. Understanding this relationship is key to solving the problem.

관련 실천

교과서 질문

FIGURE EX30.19 shows the current as a function of time through a 20-cm-long, 4.0-cm-diameter solenoid with 400 turns. Draw a graph of the induced electric field strength as a function of time at a point 1.0 cm from the axis of the solenoid.

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교과서 질문

Electricity is distributed from electrical substations to neighborhoods at 15,000 V. This is a 60 Hz oscillating (AC) voltage. Neighborhood transformers, seen on utility poles, step this voltage down to the 120 V that is delivered to your house. No energy is lost in an ideal transformer, so the output power P_out from the secondary coil equals the input power P_in to the primary coil. Suppose a neighborhood transformer delivers 250 A at 120 V. What is the current in the 15,000 V line from the substation?

171

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교과서 질문

CALC A 5.0-cm-diameter coil has 20 turns and a resistance of 0.50 Ω. A magnetic field perpendicular to the coil is B = 0.020t + 0.010t², where B is in tesla and t is in seconds. Evaluate I at t = 5 s and t = 10 s.

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교과서 질문

What is the potential difference across a 10 mH inductor if the current through the inductor drops from 150 mA to 50 mA in 10 μs? What is the direction of this potential difference? That is, does the potential increase or decrease along the direction of the current?

The magnetic field inside a 5.0-cm-diameter solenoid is 2.0 T and decreasing at 4.0 T/s. What is the electric field strength inside the solenoid at a point (a) on the axis and (b) 2.0 cm from the axis?

136

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