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During the testing of a new light bulb, a sensor is placed \(52.5 \mathrm{~cm}\) from the bulb. It records a root-mean-square value of \(9.142 \cdot 10^{-7} \mathrm{~T}\) for the magnetic field of the radiation emitted by the bulb. What is the intensity of that radiation at the sensor's location?

Short Answer

Expert verified
Based on the given information, we calculated the intensity of the radiation at the sensor's location as approximately \(1.576 * 10^{-17} W/m^2\).

Step by step solution

01

Convert the distance to meters.

The given distance is \(52.5 cm\), which can be converted to meters by dividing by 100. So, the distance in meters is: $$52.5 cm * \frac{1 m}{100 cm} =0.525 m$$
02

Calculate the Intensity using the formula.

Now we can use the formula mentioned above to calculate the intensity of the radiation at the sensor's location: Intensity = \((c * μ_0 * B_{rms}^2) / 2\) We have B_{rms}= \(9.142 * 10^{-7} T\), c=\(3 * 10^8 m/s\), and μ_0=\(4 * π * 10^{-7}T m/A\). Substitute these values into the formula and solve for Intensity: Intensity = \(\frac{(3 * 10^8\,\mathrm{m/s} * 4 * \pi *10^{-7}\,\mathrm{Tm/A} * (9.142 * 10^{-7}\,\mathrm{T})^2)}{2}\)
03

Evaluate the expression for Intensity.

Evaluating the expression for Intensity, we get: Intensity = \(\frac{(3 * 10^8\,\mathrm{m/s} * 4 * \pi *10^{-7}\,\mathrm{Tm/A} * (8.35764 * 10^{-13} \,\mathrm{T^2}))}{2}\) Intensity = \(\frac{(12 * \pi * 10^1 * 8.35764 * 10^{-20})}{2}\) Intensity = \(6 * \pi * 8.35764 * 10^{-19} W/m^2\) Intensity ≈ \(1.576 * 10^{-17} W/m^2\) So, the intensity of the radiation at the sensor's location is approximately \(1.576 * 10^{-17} W/m^2\).

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Key Concepts

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

Electromagnetic Radiation
Electromagnetic radiation encompasses a broad range of phenomena, all of which manifest as waves of electric and magnetic fields moving together through space at the speed of light, denoted by c. These waves vary in wavelength and frequency and include forms such as radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays. The common household light bulb emits visible light, which is a type of electromagnetic radiation within the specific wavelength range that our eyes can detect.

Understanding electromagnetic radiation is crucial because it includes not just the light we can see but also a host of other energy forms that we rely on daily. For instance, the radiation emitted by a light bulb enables us to see our surroundings, while radio waves transmit signals to our phones and radios.
Root-Mean-Square Value of Magnetic Field
The root-mean-square (rms) value of a magnetic field is a form of mathematical average used in physics to describe the magnitude of a varying magnetic field. If the intensity of a light bulb's electromagnetic radiation were plotted over time, it would look like a wave, with the strength of the magnetic field oscillating between positive and negative values.

In this scenario, using the rms value—typically denoted as Brms—provides a way to express the effective strength of the field for calculations like determining the radiation intensity. Despite the actual magnetic field continually changing as the waves oscillate, the Brms gives us a single measurable value to use in practical applications, such as calculating the energy conveyed by an electromagnetic wave.
Intensity of Radiation
The intensity of radiation relates to the power that radiates through a unit area and is expressed in units of watts per square meter (W/m2). It's a measure of how much energy is transmitted in the form of electromagnetic waves through a certain area over a specified period.

In our example involving a light bulb, the intensity of radiation at the location of the sensor indicates how much of the light's energy is reaching and being detected by the sensor. Calculating this intensity requires an understanding of both the nature of electromagnetic waves and the relevant physical constants—like the speed of light (c) and the permeability of free space (μ0).
Physics Problem-Solving
Physics problem-solving involves a systematic approach to understand and calculate various physical quantities. The steps to solve the problem presented here about the light bulb radiation demonstrate this approach. First, the problem is defined and relevant quantities are identified—like the sensor's distance from the light source and the Brms of the magnetic field.

Next, we apply appropriate physics formulas and constants to find the desired quantity, which in this case is the intensity of radiation. By methodically converting units, plugging in known values, and performing the arithmetic, we can derive a solution. This process showcases not just mathematical acumen, but also a deep understanding of the underlying physical principles governing how the world works.

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Most popular questions from this chapter

Two polarizing filters are crossed at \(90^{\circ}\), so when light is shined from behind the pair of filters, no light passes through. A third filter is inserted between the two, initially aligned with one of them. Describe what happens as the intermediate filter is rotated through an angle of \(360^{\circ}\)

A \(5.00-\mathrm{mW}\) laser pointer has a beam diameter of \(2.00 \mathrm{~mm}\) a) What is the root-mean-square value of the electric field in this laser beam? b) Calculate the total electromagnetic energy in \(1.00 \mathrm{~m}\) of this laser beam.

Practically everyone who has studied the electromagnetic spectrum has wondered how the world would appear if we could see over a range of frequencies comparable to the ten octaves over which we can hear rather than the less than one octave over which we can see. (An octave refers to a factor of 2 in frequency.) But this is practically impossible. Why?

Unpolarized light of intensity \(I_{0}\) is incident on a series of five polarizers, with the polarization direction of each rotated \(10.0^{\circ}\) from that of the preceding one. What fraction of the incident light will pass through the series?

The most intense beam of light that can propagate through dry air must have an electric field whose maximum amplitude is no greater than the breakdown value for air: \(E_{\max }^{\text {air }}=3.0 \cdot 10^{6} \mathrm{~V} / \mathrm{m}\), assuming that this value is unaffected by the frequency of the wave. a) Calculate the maximum amplitude the magnetic field of this wave can have. b) Calculate the intensity of this wave. c) What happens to a wave more intense than this?

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