Skip to main content
Ch 30: Electromagnetic Induction
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
All textbooksKnight Calc 5th EditionCh 30: Electromagnetic InductionProblem 83a
Chapter 30, Problem 83a

CALC Let's look at the details of eddy-current braking. A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t = 0 s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0 m. Find an expression for the loop's velocity as a function of time as it enters the magnetic field. You can ignore gravity, and you can assume that the back edge of the loop has not entered the field.

Verified step by step guidance
1
Step 1: Understand the problem setup. A square loop of side length l is moving with an initial velocity v₀ into a uniform magnetic field B. The magnetic field is perpendicular to the plane of the loop, and the loop has mass m and resistance R. We are tasked with finding the velocity of the loop as a function of time, v(t), as it enters the magnetic field. Eddy currents will be induced in the loop due to the changing magnetic flux, which will create a magnetic force opposing the motion (Lenz's Law).
Step 2: Calculate the induced EMF (electromotive force). The magnetic flux Φ through the loop is given by Φ = B × A, where A is the area of the loop inside the magnetic field. As the loop moves into the field, the area A = l × x, where x is the distance the loop has traveled into the field. Since x = v(t) × t, the flux becomes Φ = B × l × v(t) × t. The rate of change of flux induces an EMF: ε = -dΦ/dt. Using the chain rule, ε = -B × l × v(t).
Step 3: Relate the induced EMF to the current in the loop. Ohm's Law states that the current I in the loop is given by I = ε / R. Substituting the expression for ε, we get I = -(B × l × v(t)) / R.
Step 4: Determine the magnetic force acting on the loop. The magnetic force F is given by F = I × l × B (since the current flows along the sides of the loop perpendicular to the magnetic field). Substituting the expression for I, we get F = -((B² × l² × v(t)) / R). This force opposes the motion of the loop, acting as a braking force.
Step 5: Apply Newton's second law to find v(t). The net force on the loop is F = m × dv/dt. Substituting the expression for F, we get m × dv/dt = -((B² × l² × v(t)) / R). Rearrange to form a differential equation: dv/dt = -(B² × l² / (m × R)) × v(t). Solve this first-order linear differential equation to find v(t) = v₀ × exp(-t × (B² × l² / (m × R))).

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Was this helpful?

Key Concepts

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

Eddy Currents

Eddy currents are loops of electric current that are induced within conductors by a changing magnetic field due to Faraday's law of electromagnetic induction. When a conductive loop moves through a magnetic field, the change in magnetic flux through the loop generates these currents, which create their own magnetic fields opposing the motion of the loop, leading to a braking effect.
Recommended video:
Guided course
05:38
Intro to Current

Faraday's Law of Induction

Faraday's Law states that the electromotive force (EMF) induced in a circuit is directly proportional to the rate of change of magnetic flux through the circuit. This principle is fundamental in understanding how the motion of the loop in the magnetic field generates an induced EMF, which in turn produces eddy currents that affect the loop's velocity.
Recommended video:
Guided course
09:26
Faraday's Law

Magnetic Damping

Magnetic damping refers to the reduction of motion in a system due to the interaction between magnetic fields and induced currents. In the context of the square loop entering a magnetic field, the eddy currents generated create a magnetic force that opposes the loop's motion, effectively slowing it down and allowing for the analysis of its velocity as a function of time.
Recommended video:
Guided course
05:30
Magnetic Fields and Magnetic Dipoles
Related Practice
Textbook Question

CALC High-frequency signals are often transmitted along a coaxial cable, such as the one shown in FIGURE CP30.86. For example, the cable TV hookup coming into your home is a coaxial cable. The signal is carried on a wire of radius r1 while the outer conductor of radius r2 is grounded. A soft, flexible insulating material fills the space between them, and an insulating plastic coating goes around the outside. Evaluate the inductance per meter of a cable having r1 = 0.50 mm and r2 = 3.0 mm.

Textbook Question

CALC The rectangular loop in FIGURE CP30.81 has 0.020 Ω resistance. What is the induced current in the loop at this instant?

1
views
Textbook Question

CALC Let's look at the details of eddy-current braking. A square loop, length l on each side, is shot with velocity v0 into a uniform magnetic field B. The field is perpendicular to the plane of the loop. The loop has mass m and resistance R, and it enters the field at t=0 s. Assume that the loop is moving to the right along the x-axis and that the field begins at x = 0 m. Calculate and draw a graph of v over the interval 0 s ≤ t ≤ 0.04 s for the case that v0=10 m/s, l = 10 cm, m = 1.0 g, R = 0.0010 Ω, and B=0.10 T. The back edge of the loop does not reach the field during this time interval.

1
views
Textbook Question

In recent years it has been possible to buy a 1.0 F capacitor. This is an enormously large amount of capacitance. Suppose you want to build a 1.0 Hz oscillator with a 1.0 F capacitor. You have a spool of 0.25-mm-diameter wire and a 4.0-cm-diameter plastic cylinder. How long must your inductor be if you wrap it with 2 layers of closely spaced turns?

2
views
Textbook Question

The switch in FIGURE P30.77 has been open for a long time. It is closed at t = 0 s. Find an expression for the current I as a function of time. Write your expression in terms of I0, R, and L.

2
views