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. Find an expression for the induced current I(t) as a function of time.
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30. Induction and Inductance
Faraday's Law
Problem 21
Textbook Question
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 Imax 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?

1
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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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
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.
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Magnetic Field Produced by Loops and Solenoids
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.
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Intro to Electric Fields
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.
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Intro to Current
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