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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

Suppose you wanted to use Earth's magnetic field to make an ac generator at a location where the magnitude of the field is \(0.50 \mathrm{mT}\). Your coil has 1000.0 turns and a radius of \(5.0 \mathrm{cm} .\) At what angular velocity would you have to rotate it in order to generate an emf of amplitude $1.0 \mathrm{V} ?$
A coil has an inductance of \(0.15 \mathrm{H}\) and a resistance of 33 \(\Omega\). The coil is connected to a 6.0 - \(V\) ideal battery. When the current reaches half its maximum value: (a) At what rate is magnetic energy being stored in the inductor? (b) At what rate is energy being dissipated? (c) What is the total power that the battery supplies?
In Section \(20.9,\) in order to find the energy stored in an inductor, we assumed that the current was increased from zero at a constant rate. In this problem, you will prove that the energy stored in an inductor is \(U_{\mathrm{L}}=\frac{1}{2} L I^{2}-\) that is, it only depends on the current \(I\) and not on the previous time dependence of the current. (a) If the current in the inductor increases from \(i\) to \(i+\Delta i\) in a very short time \(\Delta t,\) show that the energy added to the inductor is $$\Delta U=L i \Delta i$$ [Hint: Start with \(\Delta U=P \Delta t .]\) (b) Show that, on a graph of \(L i\) versus \(i,\) for any small current interval \(\Delta i,\) the energy added to the inductor can be interpreted as the area under the graph for that interval.(c) Now show that the energy stored in the inductor when a current \(I\) flows is \(U=\frac{1}{2} L I^{2}\)
A coil of wire is connected to an ideal \(6.00-\mathrm{V}\) battery at \(t=0 .\) At \(t=10.0 \mathrm{ms},\) the current in the coil is \(204 \mathrm{mA}\) One minute later, the current is 273 mA. Find the resistance and inductance of the coil. [Hint: Sketch \(I(t) .]\)
Verify that, in SI units, \(\Delta \Phi_{\mathrm{B}} / \Delta t\) can be measured in volts - in other words, that $1 \mathrm{Wb} / \mathrm{s}=1 \mathrm{V}$
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