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 35

Chapter 29, Problem 35

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.

Verified step by step guidance

Start by understanding the concept of electromagnetic induction in solenoids. When the current in a solenoid changes, it induces an electric field around it. This is described by Faraday's law of induction.

Use the formula for the induced electric field inside a solenoid:

E = \frac{r}{R} \frac{d}{i} d t

, where

E

is the induced electric field,

r

is the distance from the axis,

R

is the radius of the solenoid, and

\frac{d}{i} d t

is the rate of change of current.

Substitute the given values into the formula:

8.00 \times 10^{- 6}

for

E

3.50

cm for

r

, and

1.10

cm for

R

. Convert these distances to meters.

Rearrange the formula to solve for

\frac{d}{i} d t

\frac{d}{i} d t

= ERr.

Calculate the value of

\frac{d}{i} d t

using the rearranged formula and the substituted values.

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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 Induction

Faraday's Law states that a changing magnetic field within a closed loop induces an electromotive force (EMF) in the loop. In the context of a solenoid, the changing current alters the magnetic field inside, inducing an electric field around the solenoid. This principle is crucial for understanding how the rate of change of current (di/dt) affects the induced electric field.

Magnetic Field Inside a Solenoid

The magnetic field inside a long, thin solenoid is uniform and 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 concept helps in understanding how the solenoid's properties and current contribute to the magnetic field, which is essential for calculating the induced electric field.

Induced Electric Field

The induced electric field around a solenoid is related to the rate of change of the magnetic field inside it. According to Faraday's Law, the magnitude of the induced electric field at a distance from the solenoid's axis can be calculated using the formula E = (r/2) * (dB/dt), where r is the radial distance from the axis. This concept is key to determining the relationship between di/dt and the given electric field.

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Related Practice

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 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 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

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?

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Textbook Question

A cardboard tube is wrapped with two windings of insulated wire wound in opposite directions, as shown in Fig. E29.20. Terminals a and b of winding A may be connected to a battery through a reversing switch. State whether the induced current in the resistor R is from left to right or from right to left in the following circumstances: (a) the current in winding Ais from a to b and is increasing; (b) the current in winding A is from b to a and is decreasing; (c) the current in winding A is from b to a and is increasing.

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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?

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.

Verified video answer for a similar problem:

Key Concepts

Faraday's Law of Induction

Magnetic Field Inside a Solenoid

Induced Electric Field

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.