Textbook Question
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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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 36: Diffraction25
- Ch 37: Special Relativity20
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
Chapter 27, Problem 12
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?
Verified step by step guidance1
First, understand that magnetic flux (Φ) through a surface is given by the formula: Φ = B * A * cos(θ), where B is the magnetic field strength, A is the area of the surface, and θ is the angle between the magnetic field and the normal (perpendicular) to the surface.
Calculate the area (A) of the rectangular surface. Convert the dimensions from centimeters to meters: 2.80 cm = 0.028 m and 3.20 cm = 0.032 m. Then, multiply these to find the area: A = 0.028 m * 0.032 m.
Identify the angle θ. Since the magnetic field is directed at an angle of 30.0° above the horizontal, the angle between the magnetic field and the normal to the surface is 90° - 30° = 60°.
Rearrange the magnetic flux formula to solve for the magnetic field strength (B): B = Φ / (A * cos(θ)).
Substitute the known values into the equation: Φ = 3.10x10^-4 Wb, A = 0.028 m * 0.032 m, and θ = 60°. Calculate cos(60°) and use these values to find B.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Magnetic Flux
Magnetic flux quantifies the amount of magnetic field passing through a given area. It is calculated as the product of the magnetic field strength, the area it penetrates, and the cosine of the angle between the field and the normal to the surface. Understanding magnetic flux is crucial for determining how the field interacts with surfaces.
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04:52
Magnetic Flux
Area Calculation
The area of a surface is essential for calculating magnetic flux. For a rectangle, the area is found by multiplying its length by its width. In this problem, the area of the rectangular surface is 2.80 cm x 3.20 cm, which must be converted to square meters for consistency in units when calculating flux.
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Trigonometric Functions
Trigonometric functions, particularly cosine, are used to resolve the component of the magnetic field perpendicular to the surface. The angle given in the problem (30.0°) is used to find the effective magnetic field component that contributes to the flux, using the formula: flux = B * A * cos(θ), where θ is the angle between the field and the surface normal.
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Related Practice
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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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?
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Textbook Question
An open plastic soda bottle with an opening diameter of 2.5 cm is placed on a table. A uniform 1.75 T magnetic field directed upward and oriented 25° from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle?
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Textbook Question
A 150 g ball containing 4.00 x 108 excess electrons is dropped into a 125 m vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 T and direction from east to west. If air resistance is negligibly small, find the magnitude and direction of the force that this magnetic field exerts on the ball just as it enters the field.
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Textbook Question
Cyclotrons are widely used in nuclear medicine for producing short-lived radioactive isotopes. These cyclotrons typically accelerate H- (the hydride ion, which has one proton and two electrons) to an energy of 5 MeV to 20 MeV. This ion has a mass very close to that of a proton because the electron mass is negligible — about 1/2000 of the proton's mass. A typical magnetic field in such cyclotrons is 1.9 T. (a) What is the speed of a 5.0 MeV H-? (b) If the H- has energy 5.0 MeV and B = 1.9 T, what is the radius of this ion's circular orbit?
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