Textbook Question
A flat, square surface with side length is in the xy-plane at . Calculate the magnitude of the flux through this surface produced by a magnetic field .
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Ch 27: Magnetic Field and Magnetic Forces
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors41
- Ch 02: Motion Along a Straight Line67
- Ch 03: Motion in Two or Three Dimensions47
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws59
- Ch 06: Work & Kinetic Energy14
- Ch 07: Potential Energy & Conservation8
- Ch 08: Momentum, Impulse, and Collisions18
- Ch 09: Rotation of Rigid Bodies42
- Ch 10: Dynamics of Rotational Motion48
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics31
- Ch 13: Gravitation35
- Ch 14: Periodic Motion32
- Ch 15: Mechanical Waves25
- Ch 16: Sound & Hearing32
- Ch 17: Temperature and Heat36
- Ch 18: Thermal Properties of Matter38
- Ch 19: The First Law of Thermodynamics33
- Ch 20: The Second Law of Thermodynamics22
- Ch 21: Electric Charge and Electric Field36
- Ch 22: Gauss' Law24
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF26
- Ch 26: Direct-Current Circuits30
- Ch 27: Magnetic Field and Magnetic Forces18
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction28
- Ch 30: Inductance17
- Ch 31: Alternating Current21
- Ch 32: Electromagnetic Waves1
- Ch 33: The Nature and Propagation of Light20
- Ch 34: Geometric Optics34
- Ch 35: Interference19
- Ch 36: Diffraction25
- Ch 37: Special Relativity20
- Ch 38: Photons: Light Waves Behaving as Particles15
- Ch 39: Particles Behaving as Waves26
- Ch 40: Quantum Mechanics I: Wave Functions21
- Ch 41: Quantum Mechanics II: Atomic Structure25
- Ch 42: Molecules and Condensed Matter18
- Ch 43: Nuclear Physics16
- Ch 44: Particle Physics and Cosmology23
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 27, Problem 5
An electron experiences a magnetic force of magnitude 4.60 x10-15 N when moving at an angle of 60.0° with respect to a magnetic field of magnitude 3.50 x 10-3 T. Find the speed of the electron.
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Identify the formula for the magnetic force on a moving charge: \( F = qvB \sin(\theta) \), where \( F \) is the magnetic force, \( q \) is the charge of the electron, \( v \) is the speed of the electron, \( B \) is the magnetic field strength, and \( \theta \) is the angle between the velocity and the magnetic field.
Rearrange the formula to solve for the speed \( v \): \( v = \frac{F}{qB \sin(\theta)} \).
Substitute the known values into the equation: \( F = 4.60 \times 10^{-15} \) N, \( q = 1.60 \times 10^{-19} \) C (the charge of an electron), \( B = 3.50 \times 10^{-3} \) T, and \( \theta = 60.0^\circ \).
Calculate \( \sin(60.0^\circ) \), which is a trigonometric function value that can be found using a calculator or trigonometric table.
Substitute all the values into the rearranged formula to find \( v \): \( v = \frac{4.60 \times 10^{-15}}{1.60 \times 10^{-19} \times 3.50 \times 10^{-3} \times \sin(60.0^\circ)} \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Magnetic Force on a Moving Charge
The magnetic force on a moving charge is given by the equation F = qvBsin(θ), where F is the force, q is the charge, v is the velocity of the charge, B is the magnetic field strength, and θ is the angle between the velocity and the magnetic field. This equation is crucial for determining the speed of the electron in the given problem.
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Trigonometric Functions
Trigonometric functions, such as sine, are used to resolve components of vectors in physics. In this context, sin(θ) is used to calculate the component of the velocity that is perpendicular to the magnetic field, which is necessary for determining the magnetic force experienced by the electron.
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Electron Charge
The charge of an electron is a fundamental constant, approximately -1.60x10^-19 C. Knowing this value is essential for calculating the speed of the electron, as it is a key variable in the equation for magnetic force. The negative sign indicates the electron's negative charge, but the magnitude is used in calculations involving force.
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Related Practice
Textbook Question
A particle with mass and a charge of has, at a given instant, a velocity . What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field ?
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Textbook Question
A horizontal rectangular surface has dimensions 2.80 cm by 3.20 cm and is in a uniform magnetic field that is directed at an angle of 30.0° above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10 x 10-4 Wb through the surface?
Textbook Question
A circular area with a radius of 6.50 cm lies in the xy-plane. What is the magnitude of the magnetic flux through this circle due to a uniform magnetic field B = 0.230 T at an angle of 53.1° from the +z-direction?