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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

An electron in the beam of a cathode-ray tube is accelerated by a potential difference of 2.00 kV. Then it passes through a region of transverse magnetic field, where it moves in a circular arc with radius 0.180 m. What is the magnitude of the field?

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?

A deuteron (the nucleus of an isotope of hydrogen) has a mass of 3.34 \(\times\) 10\(^{-27}\) kg and a charge of \(+e\). The deuteron travels in a circular path with a radius of 6.96 mm in a magnetic field with magnitude 2.50 T. (a) Find the speed of the deuteron. (b) Find the time required for it to make half a revolution. (c) Through what potential difference would the deuteron have to be accelerated to acquire this speed?

If a proton is exposed to an external magnetic field of 2 T that has a direction perpendicular to the axis of the proton's spin, what will be the torque on the proton? (a) 0; (b) 1.4 \(\times\) 10\(^{-26}\) N \(\cdot\) m; (c) 2.8 \(\times\) 10\(^{-26}\) N \(\cdot\) m; (d) 0.7 \(\times\) 10\(^{-26}\) N \(\cdot\) m.

A flat, square surface with side length 3.40 cm is in the xy-plane at \(z =\) 0. Calculate the magnitude of the flux through this surface produced by a magnetic field \(\overrightarrow{B} =\) (0.200 T)\(\hat{\imath}\) + (0.300 T)\(\hat{\jmath}\) - (0.500 T)\(\hat{k}\).

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