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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\(^\circ\) above the horizontal. What must the magnitude of the magnetic field be to produce a flux of 3.10 \(\times\) 10\(^{-4}\) Wb through the surface?

Short Answer

Expert verified
The magnetic field magnitude must be approximately 0.400 T.

Step by step solution

01

Understand the Magnetic Flux Formula

The magnetic flux \( \Phi \) through a surface is given by the formula \( \Phi = B \cdot A \cdot \cos(\theta) \), where \( B \) is the magnitude of the magnetic field, \( A \) is the area of the surface, and \( \theta \) is the angle between the magnetic field and the normal to the surface.
02

Calculate Area of the Surface

The area \( A \) of the rectangle can be calculated using the formula for the area of a rectangle: \( A = \text{length} \times \text{width} = 2.80 \text{ cm} \times 3.20 \text{ cm} = 8.96 \text{ cm}^2 \). Convert the area to square meters for consistency in units: \( 8.96 \text{ cm}^2 = 8.96 \times 10^{-4} \text{ m}^2 \).
03

Set Up the Equation for Magnetic Flux

Using the magnetic flux formula from Step 1, we have: \( 3.10 \times 10^{-4} \text{ Wb} = B \cdot 8.96 \times 10^{-4} \text{ m}^2 \cdot \cos(30.0^\circ) \).
04

Solve for Magnetic Field Magnitude \( B \)

Rearrange the magnetic flux equation to solve for \( B \): \( B = \frac{3.10 \times 10^{-4} \text{ Wb}}{8.96 \times 10^{-4} \text{ m}^2 \times \cos(30.0^\circ)} \).
05

Calculate \( \cos(30.0^\circ) \)

The value of \( \cos(30.0^\circ) \) is \( \sqrt{3}/2 \approx 0.866 \).
06

Substitute and Compute \( B \)

Substitute the values into the rearranged equation: \( B = \frac{3.10 \times 10^{-4}}{8.96 \times 10^{-4} \times 0.866} \approx 0.400 \text{ T} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Magnetic Field
A magnetic field refers to a field surrounding a magnetic material or a moving electric charge within which the force of magnetism acts. It is denoted by the symbol \( B \) and is measured in Tesla (T). Magnetic fields are fundamental to understanding the behavior of magnetic flux, which pertains to the quantity of magnetism, represented as magnetic field lines, passing through a given surface. This field can be uniform, meaning it has the same strength and direction at all points in the area it covers.

In the context of this exercise, we are dealing with a uniform magnetic field, which simplifies the calculations significantly. The strength of the magnetic field is determined, so we measure how much magnetic flux passes through the rectangular surface when the field is not perpendicular to the surface. The calculation of the magnetic flux involves the angle between the magnetic field direction and the normal (perpendicular) vector to the surface, which, in this case, is what causes the need to use the cosine of the angle. Understanding this angle's effect is crucial for effectively using trigonometric functions in the calculations.
Surface Area Calculation
Calculating surface area is essential in determining how much magnetic flux passes through a given surface. For a rectangle, such as the one in this exercise, the area \( A \) can be calculated by multiplying the length by the width of the rectangle.

Here, the rectangle measures 2.80 cm by 3.20 cm, leading to an area of \( 8.96 \text{ cm}^2 \). Since scientific calculations often require uniform units, we convert this area into square meters: \( 8.96 \times 10^{-4} \text{ m}^2 \). This standard unit ensures consistent and accurate calculations within the framework of the magnetic flux formula. Recognizing how these calculations are interconnected is key for solving problems involving magnetic fields.
Trigonometric Functions
Trigonometric functions play a vital role in calculating magnetic flux because they help determine the component of the magnetic field that is perpendicular to the surface. In this instance, the angle between the magnetic field and the surface鈥檚 normal is 30 degrees. We need to use the cosine of this angle, as the magnetic flux formula includes the term \( \cos(\theta) \), representing this perpendicular component.

The cosine of a 30-degree angle is \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \approx 0.866 \). This value is essential when plugged into the flux formula, showing how much of the magnetic field effectively penetrates the surface area.

Understanding and applying trigonometric functions accurately links directly to calculating the results in magnetic scenarios and beyond, as angles often adjust how fields impact surfaces.

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Most popular questions from this chapter

The plane of a 5.0 cm \(\times\) 8.0 cm rectangular loop of wire is parallel to a 0.19-T magnetic field. The loop carries a current of 6.2 A. (a) What torque acts on the loop? (b) What is the magnetic moment of the loop? (c) What is the maximum torque that can be obtained with the same total length of wire carrying the same current in this magnetic field?

A circular loop of wire with area \(A\) lies in the \(xy\)-plane. As viewed along the \(z\)-axis looking in the -\(z\)-direction toward the origin, a current \(I\) is circulating clockwise around the loop. The torque produced by an external magnetic field \(\overrightarrow{B}\) is given by \(\vec{\tau}\) = D(4\(\hat{\imath}\) - 3\(\hat{\jmath}\)), where \(D\) is a positive constant, and for this orientation of the loop the magnetic potential energy \(U = -\vec{\mu}\) \(\cdot\) \(\overrightarrow{B}\) is negative. The magnitude of the magnetic field is \(B_0 = 13D/IA\). (a) Determine the vector magnetic moment of the current loop. (b) Determine the components \(B_x\), \(B_y\), and \(B_z\) of \(\overrightarrow{B}\).

A group of particles is traveling in a magnetic field of unknown magnitude and direction. You observe that a proton moving at 1.50 km/s in the \(+x\)-direction experiences a force of 2.25 \(\times\) 10\(^{-16}\) N in the \(+y\)-direction, and an electron moving at 4.75 km/s in the \(-z\)-direction experiences a force of 8.50 \(\times\) 10-16 N in the \(+y\)-direction. (a) What are the magnitude and direction of the magnetic field? (b) What are the magnitude and direction of the magnetic force on an electron moving in the \(-y\)-direction at 3.20 km/s?

An electron experiences a magnetic force of magnitude 4.60 \(\times\) 10\(^{-15}\) N when moving at an angle of 60.0\(^\circ\) with respect to a magnetic field of magnitude 3.50 \(\times\) 10\(^{-3}\) T. Find the speed of the electron.

A particle with charge 7.80 \(\mu\)C is moving with velocity \(\vec{v} =\) - 13.80 \(\times\) 103m/s\(\hat{\jmath}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) (7.60 \(\times\) 10\(^{-3}\) N)\(\hat{\imath}\) - (5.20 \(\times\) 10\(^{-3}\) N)\(\hat{k}\). (a) Calculate all the components of the magnetic field you can from this information. (b) Are there components of the magnetic field that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\overrightarrow{B}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\overrightarrow{B}\) and \(\overrightarrow{F}\)?
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