Skip to main content
Ch 27: Magnetic Field and Magnetic Forces
Young & Freedman Calc - University Physics 15th Edition
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552당신이 사용하는 게 아니라요?교과서 변경
27장, 문제 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?

검증된 단계별 안내
1
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.

비슷한 문제에 대한 검증된 영상 답변:

이 영상 해법은 위 문제에 도움이 된다고 튜터들이 추천한 것입니다.
영상 길이:
3m
도움이 되었나요?

주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

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.
추천 영상:

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.
추천 영상:
가이드 코스
04:05
Calculating Work As Area Under F-x Graphs

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.
추천 영상:
가이드 코스
08:30
Intro to Wave Functions
관련 실천
교과서 질문

A flat, square surface with side length 3.40cm3.40\(\operatorname{cm}\) is in the xy-plane at z=0z = 0. Calculate the magnitude of the flux through this surface produced by a magnetic field B=(0.200T)i+(0.300T)j(0.500T)kB=(0.200T)\(\mathbf{i}\)+(0.300T)\(\mathbf{j}\)-(0.500T)\(\mathbf{k}\).

2855
views
교과서 질문

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?

3108
views
1
rank
교과서 질문

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?

2809
views
교과서 질문

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.

2089
views
교과서 질문

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 x 10-27 kg and a charge of +e. The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

2337
views
교과서 질문

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?

2396
views