Ch 29: Electromagnetic Induction

Young & Freedman Calc - University Physics 15th Edition

Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook

Young & Freedman Calc 15th Edition

Ch 29: Electromagnetic Induction

Problem 36

Chapter 29, Problem 36

A long, thin solenoid has 900 turns per meter and radius 2.50 cm. The current in the solenoid is increasing at a uniform rate of 36.0 A/s. What is the magnitude of the induced electric field at a point near the center of the solenoid and (a) 0.500 cm from the axis of the solenoid; (b) 1.00 cm from the axis of the solenoid?

Verified step by step guidance

Understand that the problem involves a solenoid with a changing current, which induces an electric field. The solenoid has 900 turns per meter, a radius of 2.50 cm, and the current is increasing at 36.0 A/s.

Recall Faraday's law of electromagnetic induction, which states that a changing magnetic field within a closed loop induces an electromotive force (EMF) in the loop. The induced electric field (E) can be found using the formula: \( E = \frac{r}{2} \frac{dB}{dt} \), where \( r \) is the distance from the axis and \( \frac{dB}{dt} \) is the rate of change of the magnetic field.

Calculate the rate of change of the magnetic field \( \frac{dB}{dt} \) inside the solenoid using the formula: \( \frac{dB}{dt} = \mu_0 n \frac{dI}{dt} \), where \( \mu_0 \) is the permeability of free space (\( 4\pi \times 10^{-7} \) T·m/A), \( n \) is the number of turns per unit length (900 turns/m), and \( \frac{dI}{dt} \) is the rate of change of current (36.0 A/s).

For part (a), substitute \( r = 0.500 \) cm (converted to meters) into the formula for the induced electric field: \( E = \frac{r}{2} \frac{dB}{dt} \). Calculate \( E \) using the previously found \( \frac{dB}{dt} \).

For part (b), repeat the process with \( r = 1.00 \) cm (converted to meters) to find the induced electric field at this distance from the axis of the solenoid.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Faraday's Law of Electromagnetic Induction

Faraday's Law states that a changing magnetic field within a closed loop induces an electromotive force (EMF) in the loop. The induced EMF is proportional to the rate of change of the magnetic flux through the loop. In the context of a solenoid, the changing current alters the magnetic field, inducing an electric field in the surrounding space.

Magnetic Field Inside a Solenoid

The magnetic field inside a long, thin solenoid is uniform and parallel to the axis of the solenoid. It is given by B = μ₀nI, where μ₀ is the permeability of free space, n is the number of turns per unit length, and I is the current. This uniform field is crucial for calculating the change in magnetic flux, which is needed to determine the induced electric field.

Induced Electric Field in a Solenoid

The induced electric field in a solenoid due to a changing current can be calculated using the relationship E = (r/2) * (dB/dt), where r is the radial distance from the axis, and dB/dt is the rate of change of the magnetic field. This formula arises from integrating the induced EMF around a circular path concentric with the solenoid, highlighting how the field varies with distance from the axis.

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Textbook Question

A metal ring 4.50 cm in diameter is placed between the north and south poles of large magnets with the plane of its area perpendicular to the magnetic field. These magnets produce an initial uniform field of 1.12 T between them but are gradually pulled apart, causing this field to remain uniform but decrease steadily at 0.250 T/s. What is the magnitude of the electric field induced in the ring?

Textbook Question

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.520 A. What is the rate at which the electric field between the plates is changing?

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Textbook Question

A long, straight solenoid with a cross-sectional area of 8.00 cm² is wound with 90 turns of wire per centimeter, and the windings carry a current of 0.350 A. A second winding of 12 turns encircles the solenoid at its center. The current in the solenoid is turned off such that the magnetic field of the solenoid becomes zero in 0.0400 s. What is the average induced emf in the second winding?

Textbook Question

A long, thin solenoid has 400 turns per meter and radius 1.10 cm. The current in the solenoid is increasing at a uniform rate di/dt. The induced electric field at a point near the center of the solenoid and 3.50 cm from its axis is 8.00 × 10^-6 V/m. Calculate di/dt.

Textbook Question

A parallel-plate, air-filled capacitor is being charged as in Fig. 29.23. The circular plates have radius 4.00 cm, and at a particular instant the conduction current in the wires is 0.520 A. (a) What is the displacement current density j_D in the air space between the plates?

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Textbook Question

The conducting rod ab shown in Fig. E29.29 makes contact with metal rails ca and db. The apparatus is in a uniform magnetic field of 0.800 T, perpendicular to the plane of the figure. In what direction does the current flow in the rod?