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Chapter 20: Problem 13

A horizontal desk surface measures \(1.3 \mathrm{m} \times 1.0 \mathrm{m} .\) If Earth's magnetic field has magnitude \(0.44 \mathrm{mT}\) and is directed \(65^{\circ}\) below the horizontal, what is the magnetic flux through the desk surface?

Short Answer

Expert verified
Question: Calculate the magnetic flux through a horizontal desk surface with dimensions \(1.3 \mathrm{m} \times 1.0 \mathrm{m}\), given Earth's magnetic field magnitude as \(0.44 \mathrm{mT}\) and the angle below the horizontal as \(65^{\circ}\). Answer: The magnetic flux through the desk surface is approximately \(4.914 \times 10^{-4}\,\mathrm{Wb}\).

Step by step solution

01

Identify the given information

We are given the following information: - Dimensions of the desk surface: \(1.3 \mathrm{m} \times 1.0 \mathrm{m}\) - Earth's magnetic field: \(0.44 \mathrm{mT} = 0.44 \times 10^{-3}\,\mathrm{T}\) - Angle below the horizontal: \(65^{\circ}\)
02

Find the component of the magnetic field that is perpendicular to the surface

The magnetic field component perpendicular to the surface can be calculated by taking the vertical component of the magnetic field. This is because the horizontal component is parallel to the surface and doesn't contribute to the magnetic flux. To find the vertical component, multiply the magnetic field magnitude by the sine of the angle below the horizontal: \(B_\perp = B \times \sin{(\theta)}\) \(B_\perp = 0.44 \times 10^{-3} \times \sin{(65^{\circ})}\) \(B_\perp \approx 0.378 \times 10^{-3}\,\mathrm{T}\)
03

Calculate the area of the desk surface

To calculate the area of the desk surface, simply multiply the length and width: \(A = 1.3 \, \mathrm{m} \times 1.0\, \mathrm{m} = 1.3\, \mathrm{m^2}\)
04

Calculate the magnetic flux through the surface

Now that we have the magnetic field component perpendicular to the desk surface and the area of the desk surface, we can calculate the magnetic flux. The magnetic flux (\(\Phi_B\)) can be calculated using the formula: \(\Phi_B = B_\mathrm{\perp} \times A\) \(\Phi_B = (0.378 \times 10^{-3}\,\mathrm{T}) \times (1.3\, \mathrm{m^2})\) \(\Phi_B \approx 4.914 \times 10^{-4}\,\mathrm{Tm^2} = 4.914 \times 10^{-4}\,\mathrm{Wb}\) So the magnetic flux through the desk surface is approximately \(4.914 \times 10^{-4}\,\mathrm{Wb}\).

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Most popular questions from this chapter

In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
An ideal inductor of inductance \(L\) is connected to an ac power supply, which provides an emf \(\mathscr{E}(t)=\mathscr{E}_{\mathrm{m}}\) sin \(\omega t\) (a) Write an expression for the current in the inductor as a function of time. [Hint: See Eq. \((20-7) .]\) (b) What is the ratio of the maximum emf to the maximum current? This ratio is called the reactance. (c) Do the maximum emf and maximum current occur at the same time? If not, how much time separates them?
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