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Ch 27: Magnetic Field and Magnetic Forces
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 27: Magnetic Field and Magnetic ForcesProblem 10
Chapter 27, Problem 10

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)\(\mathbf{k}\).

Verified step by step guidance
1
Understand the concept of magnetic flux: Magnetic flux (Φ) through a surface is given by the dot product of the magnetic field vector (B) and the area vector (A). The formula is Φ = B ⋅ A.
Determine the area vector: Since the surface is in the xy-plane, the area vector is perpendicular to the surface and points in the z-direction. The magnitude of the area vector is equal to the area of the surface, and its direction is along the positive z-axis. Therefore, A = (0, 0, A_z).
Calculate the area of the square surface: The side length of the square is 3.40 cm, which needs to be converted to meters for consistency in units. The area (A) of the square is given by A = side_length^2. Convert 3.40 cm to meters by dividing by 100.
Express the area vector: Since the surface is in the xy-plane, the area vector is A = (0, 0, A_z), where A_z is the area calculated in the previous step.
Calculate the dot product B ⋅ A: The magnetic field vector B is given as B = (0.200 T)i + (0.300 T)j - (0.500 T)k. The dot product B ⋅ A is calculated as B_x * A_x + B_y * A_y + B_z * A_z. Since A_x and A_y are zero, the dot product simplifies to B_z * A_z.

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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 through a surface is a measure of the number of magnetic field lines passing through that surface. It is calculated as the dot product of the magnetic field vector and the area vector of the surface. The formula is Φ = B · A = B * A * cos(θ), where B is the magnetic field, A is the area, and θ is the angle between B and the normal to the surface.
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Dot Product

The dot product is a mathematical operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. In the context of magnetic flux, it is used to calculate the component of the magnetic field that is perpendicular to the surface. The dot product of vectors A and B is given by A · B = Ax * Bx + Ay * By + Az * Bz.
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Area Vector

The area vector is a vector that is perpendicular to a given surface and has a magnitude equal to the area of the surface. For a flat surface in the xy-plane, the area vector points in the z-direction. In this problem, the area vector A for the square surface is (0, 0, A), where A is the area of the square, calculated as side length squared.
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Related Practice
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 particle with mass 1.81×10−31.81\(\times\)10^{-3} kg\(\operatorname{kg}\) and a charge of 1.22×10−8 C1.22\(\times\)10^{-8}\(\text{ C}\) has, at a given instant, a velocity v=(3.00×104 m/s)jv=(3.00\(\times\)10^4\(\text{ m/s}\))\(\mathbf{j}\). What are the magnitude and direction of the particle's acceleration produced by a uniform magnetic field B=(1.63T)i+(0.980T)jB=(1.63T)\(\mathbf{i}\)+(0.980T)\(\mathbf{j}\)?

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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?

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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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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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