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Chapter 27: Problem 13

An open plastic soda bottle with an opening diameter of 2.5 \(\mathrm{cm}\) is placed on a table. A uniform \(1.75-\mathrm{T}\) magnetic field directed upward and oriented \(25^{\circ}\) from vertical encompasses the bottle. What is the total magnetic flux through the plastic of the soda bottle?

Short Answer

Expert verified
The magnetic flux through the soda bottle is approximately \( 7.78 \times 10^{-4} \text{ Wb} \).

Step by step solution

01

Understand the Problem

We need to find the magnetic flux through the surface of the open plastic soda bottle. The magnetic flux is calculated as the product of the magnetic field, the area it penetrates, and the cosine of the angle between the field and the normal to the surface.
02

Calculate the Area of the Opening

The opening of the bottle is a circle with a diameter of 2.5 cm. First, we convert this diameter into meters: \[ d = 2.5 \text{ cm} = 0.025 \text{ m} \] The radius \( r \) is half of the diameter:\[ r = \frac{d}{2} = 0.0125 \text{ m} \]The area \( A \) of the circular opening is given by:\[ A = \pi r^2 = \pi (0.0125)^2 \approx 4.91 \times 10^{-4} \text{ m}^2 \]
03

Calculate the Magnetic Flux

We use the magnetic flux formula, which is:\[ \Phi = B \cdot A \cdot \cos \theta \]Where:- \( B = 1.75 \text{ T} \) is the magnetic field strength.- \( A = 4.91 \times 10^{-4} \text{ m}^2 \) is the area of the opening calculated earlier.- \( \theta = 25^{\circ} \) is the angle from vertical. This means the angle between the magnetic field and the normal to the surface (which is vertical) is \( \theta \).Substituting the values:\[ \Phi = 1.75 \cdot 4.91 \times 10^{-4} \cdot \cos(25^{\circ}) \]Calculate \( \cos(25^{\circ}) \) using a calculator:\[ \cos(25^{\circ}) \approx 0.9063 \]Thus, the magnetic flux is:\[ \Phi \approx 1.75 \cdot 4.91 \times 10^{-4} \cdot 0.9063 \approx 7.78 \times 10^{-4} \text{ Wb} \]
04

Conclusion

The total magnetic flux through the plastic of the soda bottle is approximately \( 7.78 \times 10^{-4} \text{ Wb} \).

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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 is a region around a magnetic material or a moving electric charge within which the force of magnetism acts. Magnetic fields are vector fields, usually denoted by the symbol \( B \), and are measured in teslas (T) or gauss in the CGS system. In this exercise, the magnetic field has a strength of \( 1.75 \) T, pointing upwards but slightly tilted. Understanding the direction and strength of a magnetic field is crucial when evaluating its interaction with surfaces, such as determining magnetic flux through a bottle's opening.
Area Calculation
Area calculation is an important step when determining magnetic flux through a surface. It is the surface area that the magnetic field lines penetrate. In this problem, the soda bottle鈥檚 opening is a circle. To calculate its area, you first need the diameter, which is given as \( 2.5 \) cm. Convert it into meters for consistency with SI units: \( d = 0.025 \) m.
  • Radius \( r \) is \( \frac{d}{2} \).
  • The area \( A \) of a circle is given by \( \pi r^2 \).

After doing the math, you find \( A = 4.91 \times 10^{-4} \) m\(^2\). This value is key in calculating the magnetic flux.
Cosine Function
The cosine function plays a significant role in calculating magnetic flux because it accounts for the angle between the magnetic field and the normal (perpendicular) to the surface. This exercise uses \( \cos(25^{\circ}) \), where \( 25^{\circ} \) is the given angle of inclination.
  • The cosine function ranges from -1 to 1.
  • It modifies the magnetic field's effect based on angle, reducing it if not perfectly aligned.

For \( 25^{\circ} \), \( \cos(25^{\circ}) \approx 0.9063 \), which is used in the formula:
\( \Phi = B \cdot A \cdot \cos \theta \). This function determines the proportion of the magnetic field that directly penetrates the surface.
Angle of Inclination
The angle of inclination defines the tilt between the magnetic field and the normal (perpendicular) to the surface. Here, the field is inclined at \( 25^{\circ} \) from vertical. This concept is crucial because the magnetic flux depends not just on the field strength and area, but also the field's orientation relative to the surface.
  • An angle of \( 0^{\circ} \) would mean full magnetic flux with \( \cos(0) = 1 \).
  • As the angle increases, \( \cos \theta \) decreases, reducing effective field penetration.

Understanding this angle helps predict how much of the magnetic field contributes to the actual flux through the surface.

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Most popular questions from this chapter

The Electromagnetic Pump. Magnetic forces acting on conducting fluids provide a convenient means of pumping these fluids. For example, this method can be used to pump blood without the damage to the cells that can be caused by a mechanical pump. A horizontal tube with rectangular cross section (height \(h,\) width w) is placed at right angles to a uniform magnetic field with magnitude \(B\) so that a length \(l\) is in the field (Fig. 27.75). The tube is filled with a conducting liquid, and an electric current of density \(J\) is maintained in the third mutually perpendicular direction. (a) Show that the difference of pressure between a point in the liquid on a vertical plane through \(a b\) and a point in the liquid on another vertical plane through \(c d,\) under conditions in which the liquid is prevented from flowing, is \(\Delta p=J I B\) . (b) What current density is needed to provide a pressure difference of 1.00 atm between these two points if \(B=2.20 \mathrm{T}\) and \(l=35.0 \mathrm{mm} ?\)

A particle of mass 0.195 g carries a charge of \(-2.50 \times\) \(10^{-8} \mathrm{C}\) . The particle is given an initial horizontal velocity that is due north and has magnitude \(4.00 \times 10^{4} \mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the minimum magnetic field that will keep the particle moving in the earth's gravitational field in the same horizontal, northward direction?

(a) What is the speed of a beam of electrons when the simultancous influence of an electric field of \(1.56 \times 10^{4} \mathrm{V} / \mathrm{m}\) and a magnetic field of \(4.62 \times 10^{-3} \mathrm{T}\) , with both fields normal to the beam and to each other, produces no deflection of the electrons? (b) In a diagram, show the relative orientation of the vectors \(\overrightarrow{\boldsymbol{v}}, \overrightarrow{\boldsymbol{E}},\) and \(\overrightarrow{\boldsymbol{B}} .\) (c) When the electric field is removed, what is the radius of the electron orbit? What is the period of the orbit?

A \(150-\) g ball containing \(4.00 \times 10^{8}\) excess electrons is dropped into a \(125-\mathrm{m}\) vertical shaft. At the bottom of the shaft, the ball suddenly enters a uniform horizontal magnetic field that has magnitude 0.250 \(\mathrm{T}\) 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 enrers the field.

In a 1.25 - T magnetic field directed vertically upward, a particle having a charge of magnitude 8.50\(\mu \mathrm{C}\) and initially moving northward at 4.75 \(\mathrm{km} / \mathrm{s}\) is deflected toward the east. (a) What is the sign of the charge of this particle? Make a sketch to illustrate how you found your answer. (b) Find the magnetic force on the particle.
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