Problem 52 - Physics

Question

\(E_{\mathrm{rms}}=2800 \mathrm{~N} / \mathrm{C}\). (a) What is the average intensity of the radiation? (b) The radiation is focused on a person's leg over a circular area of radius \(4.0 \mathrm{~cm} .\) What is the average power delivered to the leg? (c) The portion of the leg being radiated has a mass of \(0.28 \mathrm{~kg}\) and a specific heat capacity of \(3500 \mathrm{~J} /\left(\mathrm{kg} \cdot \mathrm{C}^{\circ}\right)\). How long does it take to raise its temperature by \(2.0 \mathrm{C}^{\circ} ?\) Assume that there is no other heat transfer into or out of the portion of the leg being heated.

Step-by-Step Solution

Answer

(a) 10.43 kW/m², (b) 52.5 W, (c) 37.3 s.

Step 1: Calculate Average Intensity

The average intensity of electromagnetic radiation can be calculated using the root mean square electric field \(E_{\mathrm{rms}}\):\[ I = \frac{c \varepsilon_0 E_{\mathrm{rms}}^2}{2} \]where \(c\) is the speed of light \( (3 \times 10^8 \text{ m/s)} \) and \(\varepsilon_0\) is the permittivity of free space \((8.85 \times 10^{-12} \text{ C}^2/(\text{N} \cdot \text{m}^2))\). Substitute the values:\[ I = \frac{(3 \times 10^8) \times (8.85 \times 10^{-12}) \times (2800)^2}{2} \]Calculating this gives:\[ I \approx 10.43 \times 10^3 \text{ W/m}^2 \]

Step 2: Calculate Power Delivered to the Leg

The power delivered can be found by multiplying the intensity \(I\) by the area \(A\). First, calculate the area of the circle (radius \(r = 4.0 \text{ cm} = 0.04 \text{ m}\)):\[ A = \pi r^2 = \pi (0.04)^2 \approx 5.03 \times 10^{-3} \text{ m}^2 \]Then, multiply by the intensity:\[ P = I \times A = 10.43 \times 10^3 \times 5.03 \times 10^{-3} \approx 52.5 \text{ W} \]

Step 3: Find Time to Raise Temperature

The energy required to raise the temperature can be calculated using the formula:\[ Q = mc\Delta T \]Substitute the mass \(m = 0.28 \text{ kg}\), specific heat capacity \(c = 3500 \text{ J/(kg} \cdot \text{C}^\circ)\), and temperature change \(\Delta T = 2.0 \text{C}^\circ\):\[ Q = 0.28 \times 3500 \times 2 = 1960 \text{ J} \]Since \(P = 52.5 \text{ W}\) is the power delivered, use \(Q = Pt\) to find the time \(t\) it takes:\[ t = \frac{Q}{P} = \frac{1960}{52.5} \approx 37.3 \text{ s} \]

Key Concepts

Average Intensity

The average intensity of electromagnetic radiation refers to the power per unit area carried by a wave. It provides a measure of how strong or weak radiation is over a given area. Knowing this helps us understand how much energy is being transferred.

It is directly linked to the root mean square (RMS) value of the electric field, a crucial aspect when dealing with electromagnetic waves. The intensity can be calculated using the formula:

• \( I = \frac{c \varepsilon_0 E_{\mathrm{rms}}^2}{2} \)
• where \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\), and \( \varepsilon_0 \) is the permittivity of free space \((8.85 \times 10^{-12} \text{ C}^2/\text{N} \, \cdot \, \text{m}^2))\).

For the given exercise with \(E_{\mathrm{rms}} = 2800 \text{ N/C}\), substituting these values into the equation determines the intensity. This is essential for understanding how radiation affects an area.

Root Mean Square Electric Field

The root mean square (RMS) electric field is a statistical measure of the magnitude of oscillating electric fields in electromagnetic waves. It provides an average value that quantifies the strength of the field over a cycle.

RMS values are crucial as they simplify the calculations involving alternating fields. Instead of dealing with varying electric fields, which would be complex, the RMS value gives a single figure to work with. The equation for the electric field portion is based on the maximum field value:

• The relation between the RMS electric field and its peak value is: \(E_{\mathrm{rms}} = \frac{E_0}{\sqrt{2}}\)
• Thus, providing a way to relate these consistent average measures to the fluctuant real-world values.

In practical cases, like the given problem, it simplifies calculations needed to determine intensity and power.

Specific Heat Capacity

Specific heat capacity characterizes how a substance absorbs heat energy. It describes the amount of heat required to change the temperature of a unit mass of a substance by one degree Celsius. In terms of units, it is often expressed as \( \text{J/(kg} \, \cdot \, \text{C}^\circ)\).

This concept is vital for determining how an object absorbs and disperses heat energy. In the exercise, it refers to the mass of a leg absorbing radiant energy to produce a temperature increase. The formula used is:

• \( Q = mc\Delta T \)
• where \( m \) is mass, \( c \) is specific heat capacity, and \( \Delta T \) is the temperature change.

Calculating \( Q \), the heat energy absorbed, allows us to relate the thermal properties of matter to electromagnetic energy transfer.

Power Calculation

Power in the context of electromagnetic radiation refers to the rate at which energy is transferred or converted. It is expressed in watts (W). When dealing with radiation, power calculation helps to quantify the energy interaction with surfaces.

From intensity, we calculate power by considering the area affected. The power delivered onto a surface is given by:

• \( P = I \cdot A \)
• where \( I \) is intensity and \( A \) is area.

Combining this with specific heat capacity calculations provides insights into how long a medium retains or transfers energy. It represents energy efficiency and time dependency within electromagnetic interactions, acting as a bridge between raw energy input and thermal effects.