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Two flat surfaces are exposed to a uniform, horizontal magnetic field of magnitude 0.47 T. When viewed edge-on, the first surface is tilted at an angle of \(12^{\circ}\) from the horizontal, and a net magnetic flux of \(8.4 \times 10^{-3} \mathrm{Wb}\) passes through it. The same net magnetic flux passes through the second surface. (a) Determine the area of the first surface. (b) Find the smallest possible value for the area of the second surface.

Short Answer

Expert verified
(a) The area of the first surface is approximately 0.0182 m². (b) The smallest area of the second surface is approximately 0.0179 m².

Step by step solution

01

Understanding Magnetic Flux Formula

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

Solving for the Area of the First Surface (a)

The problem states that \( \Phi = 8.4 \times 10^{-3} \text{ Wb} \), \( B = 0.47 \text{ T} \), and \( \theta = 12^\circ \). We can rearrange the magnetic flux formula to solve for \( A \):\[ A = \frac{\Phi}{B \cos(\theta)} \]Substitute the known values:\[ A = \frac{8.4 \times 10^{-3}}{0.47 \cos(12^\circ)} \]Calculate \( \cos(12^\circ) \approx 0.9781 \), then:\[ A \approx \frac{8.4 \times 10^{-3}}{0.47 \times 0.9781} \approx 0.0182 \text{ m}^2 \]
03

Introduction to the Second Surface Problem (b)

To find the smallest area for the second surface, the angle \( \theta \) must be such that \( \cos(\theta) \) is largest, ideally 1. This occurs when the surface is perpendicular to the magnetic field.
04

Calculating Minimum Area for Surface 2 (b)

Using the perpendicular condition, \( \cos(0^\circ) = 1 \). The area \( A_2 \) for the second surface is:\[ A_2 = \frac{\Phi}{B} \]Substitute the known values:\[ A_2 = \frac{8.4 \times 10^{-3}}{0.47} \approx 0.0179 \text{ m}^2 \]

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Key Concepts

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

Magnetic Field Strength
In the study of magnetic fields, magnetic field strength is a core concept. It is denoted by the symbol \( B \) and measured in Tesla (T). Magnetic field strength quantifies the force and intensity of the magnetic field lines. For this problem, the field has a strength of 0.47 T.

Understanding magnetic field strength is essential because it plays a critical role in determining magnetic flux. The stronger the field, the greater the flux passing through a given area if other factors remain constant.

Consider magnetic fields like invisible lines passing through space. The field strength tells us how "crowded" these lines are in any area. In simpler terms, more lines mean a stronger field and more interactions with surfaces they encounter.
Angle of Inclination
The angle of inclination, denoted as \( \theta \), is the angle between the magnetic field direction and the line perpendicular to the surface. In this scenario, it is given as \( 12^{\circ} \) for the first surface.

This angle is crucial in calculating magnetic flux because it affects how much of the field effectively passes through the surface. The function \( \cos(\theta) \) helps determine how much of the magnetic field is effectively "seen" by the surface.

For instance, if a surface is perfectly perpendicular to the field (\( \theta = 0^{\circ} \)), the entire field strength affects the flux. As the angle increases, less of the field contributes to the flux. Thus, angles closer to zero maximize the flux for a defined area and field strength.
Surface Area Calculation
Calculating surface area \( A \) is essential to determine the magnetic flux through a surface. Given in the formula \( \Phi = B A \cos(\theta) \), solving for \( A \) when \( \Phi \), \( B \), and \( \theta \) are known involves rearranging the formula.

The formula reconfiguration is \[ A = \frac{\Phi}{B \cos(\theta)} \], where the cosine adjustment accounts for the inclination of the surface relative to the field. This adjustment ensures that we measure only the effective area contributing to the flux, not just the total physical area of the surface.

For our problem, calculating \( A \) for the first surface entails plugging in the values for \( \Phi \), \( B \), and \( \cos(12^{\circ}) \), resulting in \( A \approx 0.0182 \text{ m}^2 \). This represents the portion of the surface "seen" by the magnetic field strength due to its inclination.
Perpendicular Magnetic Field
A perpendicular magnetic field occurs when the surface is aligned such that the magnetic lines strike it at a right angle, meaning \( \theta = 0^{\circ} \). In this orientation, \( \cos(\theta) = 1 \), maximizing the magnetic flux through the surface.

For the second surface in our problem, achieving perpendicular alignment allows us to calculate the smallest possible area \( A_2 \) while maintaining the same magnetic flux. With \( \cos(0^{\circ}) = 1 \), the formula simplifies to \( A_2 = \frac{\Phi}{B} \).

By plugging in the values for \( \Phi \) and \( B \), we find \( A_2 \approx 0.0179 \text{ m}^2 \). This reduced area highlights the importance of aligning surfaces perpendicularly to magnetic fields for efficient flux penetration.

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Most popular questions from this chapter

A flat circular coil with 105 turns, a radius of \(4.00 \times 10^{-2} \mathrm{m}\), and a resistance of \(0.480 \Omega\) is exposed to an external magnetic field that is directed perpendicular to the plane of the coil. The magnitude of the external magnetic field is changing at a rate of \(\Delta B / \Delta t=0.783 \mathrm{T} / \mathrm{s}\), thereby inducing a current in the coil. Find the magnitude of the magnetic field at the center of the coil that is produced by the induced current.

A motor is designed to operate on \(117 \mathrm{V}\) and draws a current of \(12.2 \mathrm{A}\) when \(\mathrm{it}\) first starts up. At its normal operating speed, the motor draws a current of 2.30 A. Obtain (a) the resistance of the armature coil, (b) the back emf developed at normal speed, and (c) the current drawn by the motor at one-third of the normal speed.

A circular coil of radius 0.11 \(\mathrm{m}\) contains a single turn and is located in a constant magnetic field of magnitude \(0.27 \mathrm{T}\). The magnetic field has the same direction as the normal to the plane of the coil. The radius increases to \(0.30 \mathrm{m}\) in a time of \(0.080 \mathrm{s}\). Concepts: (i) Why is there an emf induced in the coil? (ii) Does the magnitude of the induced emf depend on whether the area is increasing or decreasing? Explain. (iii) What determines the amount of current induced in the coil? (iv) If the coil is cut so it is no longer one continuous piece, are there an induced emf and an induced current? Explain. Calculations: (a) Determine the magnitude of the emf induced in the coil. (b) The coil has a resistance of \(0.70 \Omega .\) Find the magnitude of the induced current.

A circular coil (950 turns, radius \(=0.060 \mathrm{m}\) ) is rotating in a uniform magnetic field. At \(t=0\) s, the normal to the coil is perpendicular to the magnetic field. At \(t=0.010 \mathrm{s}\), the normal makes an angle of \(\phi=45^{\circ}\) with the field because the coil has made one-eighth of a revolution. An average emf of magnitude \(0.065 \mathrm{V}\) is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

A \(3.0-\mu \mathrm{F}\) capacitor has a voltage of \(35 \mathrm{V}\) between its plates. What must be the current in a \(5.0-\mathrm{mH}\) inductor so that the energy stored in the inductor equals the energy stored in the capacitor?

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