Skip to main content
Ch 31: Electromagnetic Fields and Waves
Knight Calc - Physics for Scientists and Engineers 5th Edition
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoKnight Calc 5th EditionCh 31: Electromagnetic Fields and WavesProblema 28
Capitolo 31, Problema 28

What is the force (magnitude and direction) on the proton in FIGURE P31.28? Give the direction as an angle cw or ccw from vertical.

Guida verificata passo dopo passo
1
Step 1: Identify the forces acting on the proton. The proton experiences two forces: the electric force due to the electric field \( E \) and the magnetic force due to the magnetic field \( B \). The electric force is given by \( F_E = qE \), and the magnetic force is given by \( F_B = qvB \sin \theta \), where \( \theta \) is the angle between the velocity \( \vec{v} \) and the magnetic field \( \vec{B} \).
Step 2: Calculate the electric force. The charge of the proton is \( q = 1.6 \times 10^{-19} \, \text{C} \), and the electric field \( E \) is \( 1.0 \times 10^6 \, \text{V/m} \). Use the formula \( F_E = qE \) to find the magnitude of the electric force.
Step 3: Calculate the magnetic force. The proton's velocity \( v \) is \( 1.0 \times 10^7 \, \text{m/s} \), the magnetic field \( B \) is \( 0.10 \, \text{T} \), and the angle \( \theta \) between \( \vec{v} \) and \( \vec{B} \) is \( 90^\circ \) (since \( \vec{v} \) is horizontal and \( \vec{B} \) is perpendicular to the plane). Use the formula \( F_B = qvB \sin \theta \) to find the magnitude of the magnetic force.
Step 4: Determine the net force. The electric force \( \vec{F_E} \) acts vertically upward, while the magnetic force \( \vec{F_B} \) acts horizontally to the left. Use vector addition to find the magnitude and direction of the net force. The magnitude is \( F_{\text{net}} = \sqrt{F_E^2 + F_B^2} \), and the direction is given by \( \tan \phi = \frac{F_B}{F_E} \), where \( \phi \) is the angle of the net force relative to the vertical.
Step 5: Express the direction of the net force. Convert the angle \( \phi \) into a clockwise (cw) or counterclockwise (ccw) direction from the vertical, depending on the orientation of the forces. The magnetic force is to the left, and the electric force is upward, so the net force will be at an angle \( \phi \) ccw from the vertical.

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Electric Field (E)

An electric field is a region around a charged particle where other charged particles experience a force. It is quantified in volts per meter (V/m) and indicates the direction a positive test charge would move. In this scenario, the electric field strength is given as 1.0 × 10^6 V/m, which influences the proton's motion.
Video consigliato:
Percorso guidato
03:16
Intro to Electric Fields

Magnetic Field (B)

A magnetic field is a vector field that exerts a force on moving charges and magnetic dipoles. It is measured in teslas (T) and can affect the trajectory of charged particles. In this case, the magnetic field strength is 0.10 T, which will interact with the proton's velocity to determine the net force acting on it.
Video consigliato:
Percorso guidato
05:30
Magnetic Fields and Magnetic Dipoles

Lorentz Force

The Lorentz force is the combined force exerted on a charged particle due to electric and magnetic fields. It is calculated using the formula F = q(E + v × B), where q is the charge, E is the electric field, v is the velocity, and B is the magnetic field. This force will dictate both the magnitude and direction of the force acting on the proton in the given scenario.
Video consigliato:
Percorso guidato
13:39
Lorentz Transformations of Velocity
Pratica correlata
Domanda del libro di testo

Unpolarized light with intensity 350 W/m2 passes first through a polarizing filter with its axis vertical, then through a second polarizing filter. It emerges from the second filter with intensity 131 W/m2. What is the angle from vertical of the axis of the second polarizing filter?

186
views
Domanda del libro di testo

In FIGURE P31.32, a circular loop of radius r travels with speed v along a charged wire having linear charge density λ. The wire is at rest in the laboratory frame, and it passes through the center of the loop. What electric and magnetic fields would an experimenter in the loop's frame calculate at distance r from the current of part c?

139
views
Domanda del libro di testo

FIGURE EX31.23 shows a horizontally polarized radio wave of frequency 1.0×106 Hz traveling into the figure. The maximum electric field strength is 1000 V/m. What is the maximum magnetic field strength?

88
views
Domanda del libro di testo

A proton is fired with a speed of 1.0×106 m/s through the parallel-plate capacitor shown in FIGURE P31.29. The capacitor's electric field is E =(1.0×105 V/m, down). How does an experimenter in the proton's frame explain that the proton experiences no force as the charged plates fly by?

348
views
Domanda del libro di testo

An electron travels with v⃗=(5.0×106i^)m/s\(\vec{v}\) = (5.0 \(\times\) 10^6\,\(\hat{i}\))\,\(\text{m/s}\) through a point in space where E⃗=(2.0×105i^−2.0×105j^)V/m\(\vec{E}\) = (2.0 \(\times\) 10^5\,\(\hat{i}\) - 2.0 \(\times\) 10^5\,\(\hat{j}\))\,\(\text{V/m}\) and B⃗=−0.10k^T\(\vec{B}\) = -0.10\,\(\hat{k}\)\,\(\text{T}\). What is the force on the electron?

304
views
Domanda del libro di testo

A microwave beam with a wavelength of 1.5 cm has an intensity of 25 W/m2. What is the magnetic field amplitude?

80
views