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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

In this problem, you derive the expression for the self inductance of a long solenoid [Eq. \((20-15 a)] .\) The solenoid has \(n\) turns per unit length, length \(\ell,\) and radius \(r\) Assume that the current flowing in the solenoid is \(I\) (a) Write an expression for the magnetic field inside the solenoid in terms of \(n, \ell, r, I,\) and universal constants. (b) Assume that all of the field lines cut through each turn of the solenoid. In other words, assume the field is uniform right out to the ends of the solenoid-a good approximation if the solenoid is tightly wound and sufficiently long. Write an expression for the magnetic flux through one turn. (c) What is the total flux linkage through all turns of the solenoid? (d) Use the definition of self-inductance [Eq. \((20-14)]\) to find the self inductance of the solenoid.
The current in a \(0.080-\mathrm{H}\) solenoid increases from \(20.0 \mathrm{mA}\) to \(160.0 \mathrm{mA}\) in \(7.0 \mathrm{s} .\) Find the average emf in the solenoid during that time interval.
A 2 -m-long copper pipe is held vertically. When a marble is dropped down the pipe, it falls through in about 0.7 s. A magnet of similar size and shape takes much longer to fall through the pipe. (a) As the magnet is falling through the pipe with its north pole below its south pole, what direction do currents flow around the pipe above the magnet? Below the magnet (CW or CCW as viewed from the top)? (b) Sketch a graph of the speed of the magnet as a function of time. [Hint: What would the graph look like for a marble falling through honey?]
The component of the external magnetic field along the central axis of a 50 -turn coil of radius \(5.0 \mathrm{cm}\) increases from 0 to 1.8 T in 3.6 s. (a) If the resistance of the coil is \(2.8 \Omega,\) what is the magnitude of the induced current in the coil? (b) What is the direction of the current if the axial component of the field points away from the viewer?
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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