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A long solenoid of length \(8.0 \times 10^{-2} \mathrm{m}\) and cross-sectional area \(5.0 \times 10^{-5} \mathrm{m}^{2}\) contains 6500 turns per meter of length. Determine the emf induced in the solenoid when the current in the solenoid changes from 0 to 1.5 A during the time interval from 0 to 0.20 s.

Short Answer

Expert verified
The induced EMF in the solenoid is approximately \( 1.599 \text{ mV} \).

Step by step solution

01

Know the Formula for EMF Induction

The formula for the electromotive force (EMF) induced in a solenoid is given by Faraday's Law: \( \text{EMF} = -N \frac{d\Phi}{dt} \), where \( N \) is the total number of turns and \( \Phi \) is the magnetic flux.
02

Calculate the Total Number of Turns

The total number of turns \( N \) in the solenoid is found using the formula: \( N = n \cdot L \), where \( n = 6500 \) turns/m, and \( L = 8.0 \times 10^{-2} \text{ m} \). So, \( N = 6500 \times 8.0 \times 10^{-2} = 520 \) turns.
03

Determine the Change in Magnetic Flux

The magnetic flux \( \Phi \) through the cross-section of the solenoid is given by \( \Phi = B \cdot A \), where \( B \) is the magnetic field and \( A = 5.0 \times 10^{-5} \text{ m}^2 \). The magnetic field inside a solenoid is \( B = \mu_0 \cdot n \cdot I \). Therefore, the change in flux \( \Delta \Phi = (B_f - B_i) \cdot A \), where \( B_f = \mu_0 \cdot n \cdot I_f \), and \( B_i = \mu_0 \cdot n \cdot I_i \). Given \( I_i = 0 \text{ A} \) and \( I_f = 1.5 \text{ A} \).
04

Calculate the Change in Magnetic Field

\( B_f = \mu_0 \cdot 6500 \cdot 1.5 \), where \( \mu_0 = 4\pi \times 10^{-7} \text{ Tm/A} \). Therefore, \( B_f = 4\pi \times 10^{-7} \times 6500 \times 1.5 = 1.23 \times 10^{-2} \text{ T} \). Since \( B_i = 0 \) because initial current \( I_i = 0 \), \( \Delta B = 1.23 \times 10^{-2} \text{ T} \).
05

Calculate the Change in Flux

The change in magnetic flux \( \Delta \Phi = \Delta B \cdot A = 1.23 \times 10^{-2} \times 5.0 \times 10^{-5} = 6.15 \times 10^{-7} \text{ Wb} \).
06

Compute the EMF

Using Faraday's law, \( \text{EMF} = -N \frac{\Delta \Phi}{\Delta t} \), where \( \Delta t = 0.20 \text{ s} \), \( \text{EMF} = -520 \frac{6.15 \times 10^{-7}}{0.20} = -1.599 \times 10^{-3} \text{ V} \).
07

Significance of the Negative Sign

The negative sign in the formula represents Lenz's law, indicating that the induced EMF will oppose the change in current. Hence, the magnitude of the EMF is \( 1.599 \text{ mV} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Faraday's Law
Understanding Faraday's Law is essential when studying electromagnetic induction. This fundamental principle explains how changing magnetic fields can induce an electromotive force (EMF) in a circuit. According to Faraday's Law, the induced EMF in a closed circuit is proportional to the rate of change of the magnetic flux through the circuit. The mathematical expression for this is given by:
  • \( \text{EMF} = -N \frac{d\Phi}{dt} \)
Here, \( N \) is the number of turns of the coil, while \( \Phi \) represents the magnetic flux. The negative sign shows that the induced EMF opposes the change in flux, a concept known as Lenz's Law. To fully grasp this law, consider how a solenoid acts when the current through it changes, affecting the magnetic flux and subsequently inducing EMF.
Magnetic Flux
Magnetic flux (\( \Phi \)) plays a crucial role in the context of electromagnetic induction. It is defined as the product of the magnetic field (\( B \)) and the area (\( A \)) it penetrates perpendicularly. Mathematically, flux can be expressed as:
  • \( \Phi = B \cdot A \)
In a solenoid, the magnetic field is uniform and can be related to the current and number of turns per unit length as \( B = \mu_0 n I \), where \( \mu_0 \) is the permeability of free space and \( n \) is the number of turns per unit length. When the current changes, it leads to a change in magnetic flux (\( \Delta \Phi \)), which is crucial in calculating the induced EMF using Faraday's Law. This relationship demonstrates how intimately connected magnetic flux and EMF are, especially in devices like solenoids.
Solenoid
A solenoid is a long coil of wire wrapped in many turns, used to create a uniform magnetic field when an electric current flows through it. It is a key component in studies involving electromagnetic induction. Solenoids work by:
  • Producing a magnetic field when electrical current flows through.
  • Enhancing the magnetic field due to the many windings of the coil.
The strength of the magnetic field inside a solenoid depends on the current intensity and the number of turns per unit length. Equation \( B = \mu_0 n I \) shows this dependence, where \( I \) is the current. In the exercise given, the solenoid’s dimensions and current change lead us to calculate the flux variation and consequently, the induced EMF. This happens as the magnetic field inside the solenoid reacts to the increase in current, demonstrating real-life applications of Faraday's and Lenz's Laws.
Lenz's Law
Lenz's Law serves a critical role in explaining the direction of the induced electromotive force (EMF). It states that the induced EMF will always oppose the change in magnetic flux that produced it. This opposition is represented as the negative sign in Faraday's Law:
  • \( \text{EMF} = -N \frac{d\Phi}{dt} \)
Lenz's Law ensures that the energy conservation principle holds in electromagnetic systems. In practice, this means that if the magnetic field through a solenoid is increased, the induced EMF generates a current opposing this increase. It’s like nature's way of maintaining balance. When analyzing the solenoid problem, the resulting negative sign in the EMF calculation is a direct consequence of Lenz's Law, indicating the induced EMF’s direction is opposite to the change that prompted it. This concept helps us understand how circuits react to dynamic magnetic environments.

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Most popular questions from this chapter

Two \(0.68-\mathrm{m}\) -long conducting rods are rotating at the same speed in opposite directions, and both are perpendicular to a 4.7-T magnetic field. As the drawing shows, the ends of these rods come to within \(1.0 \mathrm{mm}\) of each other as they rotate. Moreover, the fixed ends about which the rods are rotating are connected by a wire, so these ends are at the same electric potential. If a potential difference of \(4.5 \times 10^{3} \mathrm{V}\) is required to cause a 1.0 -mm spark in air, what is the angular speed (in \(\mathrm{rad} / \mathrm{s}\) ) of the rods when a spark jumps across the gap?

A flat coil of wire has an area \(A, N\) turns, and a resistance \(R\). It is situated in a magnetic field, such that the normal to the coil is parallel to the magnetic field. The coil is then rotated through an angle of \(90^{\circ},\) so that the normal becomes perpendicular to the magnetic field. The coil has an area of \(1.5 \times 10^{-3} \mathrm{m}^{2}, 50\) turns, and a resistance of \(140 \Omega .\) During the time while it is rotating, a charge of \(8.5 \times 10^{-5} \mathrm{C}\) flows in the coil. What is the magnitude of the magnetic field?

Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of \(12^{\circ}\) from the horizontal, and a net magnetic flux of \(8.4 \times 10^{-3} \mathrm{Wb}\) passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

The battery charger for an MP3 player contains a step-down transformer with a turns ratio of \(1: 32,\) so that the voltage of \(120 \mathrm{V}\) available at a wall socket can be used to charge the battery pack or operate the player. What voltage does the secondary coil of the transformer provide?

A solenoid has a cross-sectional area of \(6.0 \times 10^{-4} \mathrm{m}^{2},\) consists of 400 turns per meter, and carries a current of 0.40 A. A 10 -turn coil is wrapped tightly around the circumference of the solenoid. The ends of the coil are connected to a \(1.5-\Omega\) resistor. Suddenly, a switch is opened, and the current in the solenoid dies to zero in a time of 0.050 s. Find the average current induced in the coil.

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