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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’s 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

A dc motor with its rotor and field coils connected in series has an internal resistance of 3.2 \(\Omega\). When the motor is running at full load on a 120-V line, the emf in the rotor is 105 V. (a) What is the current drawn by the motor from the line? (b) What is the power delivered to the motor? (c) What is the mechanical power developed by the motor?

A particle with charge -5.60 nC is moving in a uniform magnetic field \(\overrightarrow{B} =\) -(1.25 T)\(\hat{k}\). The magnetic force on the particle is measured to be \(\overrightarrow{F} =\) -(3.40 \(\times\) 10\(^{-7}\)N)\(\hat{\imath}\) + (7.40 \(\times\) 10\(^{-7}\)N)\(\hat{\jmath}\). (a) Calculate all the components of the velocity of the particle that you can from this information. (b) Are there components of the velocity that are not determined by the measurement of the force? Explain. (c) Calculate the scalar product \(\vec{v}\) \(\cdot\) \(\overrightarrow{F}\). What is the angle between \(\vec{v}\) and \(\overrightarrow{F}\)?

The amount of meat in prehistoric diets can be determined by measuring the ratio of the isotopes \(^{15}\)N to \(^{14}\)N in bone from human remains. Carnivores concentrate \(^{15}\)N, so this ratio tells archaeologists how much meat was consumed. For a mass spectrometer that has a path radius of 12.5 cm for \(^{12}\)C ions (mass 1.99 \(\times\) 10\(^{-26}\) kg), find the separation of the \(^{14}\)N 1mass 2.32 \(\times\) 10\(^{-26}\) kg2 and 15N (mass 2.49 \(\times\) 10\(^{-26}\) kg) isotopes at the detector.

Singly ionized (one electron removed) atoms are accelerated and then passed through a velocity selector consisting of perpendicular electric and magnetic fields. The electric field is 155 V/m and the magnetic field is 0.0315 T. The ions next enter a uniform magnetic field of magnitude 0.0175 T that is oriented perpendicular to their velocity. (a) How fast are the ions moving when they emerge from the velocity selector? (b) If the radius of the path of the ions in the second magnetic field is 17.5 cm, what is their mass?

A singly ionized (one electron removed) \(^{40}\)K atom passes through a velocity selector consisting of uniform perpendicular electric and magnetic fields. The selector is adjusted to allow ions having a speed of 4.50 km/s to pass through undeflected when the magnetic field is 0.0250 T. The ions next enter a second uniform magnetic field (\(B'\)) oriented at right angles to their velocity. \(^{40}\)K contains 19 protons and 21 neutrons and has a mass of 6.64 \(\times\) 10\(^{-26}\) kg. (a) What is the magnitude of the electric field in the velocity selector? (b) What must be the magnitude of \(B'\) so that the ions will be bent into a semicircle of radius 12.5 cm?

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