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Chapter 20: Problem 78

The largest constant magnetic field achieved in the laboratory is about $40 \mathrm{T}$. (a) What is the magnetic energy density due to this field? (b) What magnitude electric field would have an equal energy density?

Short Answer

Expert verified
Answer: The magnetic energy density of the given magnetic field is approximately \(3.2 \pi 脳 10^{-4} \frac{J}{m^3}\), and the magnitude of an electric field with equal energy density is approximately \(9.45 脳 10^6 V/m\).

Step by step solution

01

Calculate the magnetic energy density

First, we need to find the magnetic energy density using the magnetic field value. The formula for magnetic energy density (u) is: $$ u = \frac{1}{2} \mu_0 B^2 $$ where \(\mu_0 = 4\pi 脳 10^{-7} T m/A\) is the permeability of free space, and \(B = 40 T\) is the magnetic field. Plug in the values to get the magnetic energy density: $$ u = \frac{1}{2} 脳 (4\pi 脳 10^{-7} T m/A) 脳 (40 T)^2 $$
02

Solve for the magnetic energy density

Calculate the magnetic energy density: $$ u = \frac{1}{2} 脳 (4\pi 脳 10^{-7} T m/A) 脳 (1600 T^2) $$ $$ u = 3.2 \pi 脳 10^{-4} \frac{J}{m^3} $$ Now we have the magnetic energy density due to the magnetic field.
03

Calculate the electric field with equal energy density

Next, we want to find the magnitude of an electric field with the same energy density. The formula for electric energy density (u) is: $$ u = \frac{1}{2} \epsilon_0 E^2 $$ where \(\epsilon_0 = 8.85 脳 10^{-12} F/m\) is the permittivity of free space, and \(E\) is the electric field. Since the electric energy density is equal to the magnetic energy density, we can set the two formulas equal to each other: $$ \frac{1}{2} \mu_0 B^2 = \frac{1}{2} \epsilon_0 E^2 $$ Now we need to solve for the electric field E: $$ \frac{\mu_0 B^2}{\epsilon_0} = E^2 $$
04

Solve for the electric field

Plug in the values for \(\mu_0\), \(B\), and \(\epsilon_0\): $$ \frac{(4\pi 脳 10^{-7} T m/A) 脳 (40 T)^2}{8.85 脳 10^{-12} F/m} = E^2 $$ Calculate the electric field and take the square root of the result: $$ E = \sqrt{\frac{(4\pi 脳 10^{-7} T m/A) 脳 (1600 T^2)}{8.85 脳 10^{-12} F/m}} $$ $$ E 鈮� 9.45 脳 10^6 V/m $$ The electric field of magnitude \(9.45 脳 10^6 V/m\) has an equal energy density to the given magnetic field.

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Most popular questions from this chapter

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?
A \(0.30-\mathrm{H}\) inductor and a \(200.0-\Omega\) resistor are connected in series to a \(9.0-\mathrm{V}\) battery. (a) What is the maximum current that flows in the circuit? (b) How long after connecting the battery does the current reach half its maximum value? (c) When the current is half its maximum value, find the energy stored in the inductor, the rate at which energy is being stored in the inductor, and the rate at which energy is dissipated in the resistor. (d) Redo parts (a) and (b) if, instead of being negligibly small, the internal resistances of the inductor and battery are \(75 \Omega\) and \(20.0 \Omega\) respectively.
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