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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 \(3.2 \pi × 10^{-4} \frac{J}{m^3}\), and the magnitude of an electric field with equal energy density is approximately \(9.45 × 10^6 V/m\).

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 \(\mu_0 = 4\pi × 10^{-7} T m/A\) is the permeability of free space, and \(B = 40 T\) is the magnetic field. Plug in the values to get the magnetic energy density: $$ u = \frac{1}{2} × (4\pi × 10^{-7} T m/A) × (40 T)^2 $$
02

Solve for the magnetic energy density

Calculate the magnetic energy density: $$ u = \frac{1}{2} × (4\pi × 10^{-7} T m/A) × (1600 T^2) $$ $$ u = 3.2 \pi × 10^{-4} \frac{J}{m^3} $$ 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 \(\epsilon_0 = 8.85 × 10^{-12} F/m\) 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 \(\epsilon_0\): $$ \frac{(4\pi × 10^{-7} T m/A) × (40 T)^2}{8.85 × 10^{-12} F/m} = E^2 $$ Calculate the electric field and take the square root of the result: $$ E = \sqrt{\frac{(4\pi × 10^{-7} T m/A) × (1600 T^2)}{8.85 × 10^{-12} F/m}} $$ $$ E ≈ 9.45 × 10^6 V/m $$ The electric field of magnitude \(9.45 × 10^6 V/m\) has an equal energy density to the given magnetic field.

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Most popular questions from this chapter

Tim is using a cordless electric weed trimmer with a de motor to cut the long weeds in his back yard. The trimmer generates a back emf of \(18.00 \mathrm{V}\) when it is connected to an emf of \(24.0 \mathrm{V}\) de. The total electrical resistance of the electric motor is \(8.00 \Omega .\) (a) How much current flows through the motor when it is running smoothly? (b) Suddenly the string of the trimmer gets wrapped around a pole in the ground and the motor quits spinning. What is the current through the motor when there is no back emf? What should Tim do?
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