/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 68 A magnetic field has a magnitude... [FREE SOLUTION] | 91Ó°ÊÓ

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A magnetic field has a magnitude of \(12 \mathrm{~T}\). What is the magnitude of an electric field that stores the same energy per unit volume as this magnetic field?

Short Answer

Expert verified
The electric field magnitude is approximately 3.42 MV/m.

Step by step solution

01

Understand the formula for energy density in a magnetic field

The energy density for a magnetic field can be described by the formula \( u_B = \frac{B^2}{2\mu_0} \), where \( u_B \) is the energy density, \( B \) is the magnetic field strength, and \( \mu_0 \) is the permeability of free space, equal to \( 4\pi \times 10^{-7} \).
02

Calculate the energy density of the magnetic field

Substitute the given magnetic field magnitude \( B = 12 \mathrm{~T} \) into the energy density formula: \[ u_B = \frac{(12)^2}{2 \times 4\pi \times 10^{-7}} \]. Calculate and simplify to get the energy density.
03

Understand the formula for energy density in an electric field

The energy density for an electric field is given by \( u_E = \frac{\varepsilon_0 E^2}{2} \), where \( u_E \) is the energy density, \( E \) is the electric field strength, and \( \varepsilon_0 \) is the permittivity of free space, equal to \( 8.854 \times 10^{-12} \).
04

Equate the energy densities of the magnetic and electric fields

Set the magnetic energy density equal to the electric field energy density: \( u_B = u_E \). This leads to \( \frac{B^2}{2\mu_0} = \frac{\varepsilon_0 E^2}{2} \). Simplify this equation to solve for \( E \).
05

Solve for the electric field magnitude

Rearrange the equation from the previous step to solve for \( E \): \[ E = \sqrt{\frac{B^2}{\mu_0 \varepsilon_0}} \], substitute known values and solve: \( E = \sqrt{\frac{(12)^2}{4\pi \times 10^{-7} \times 8.854 \times 10^{-12}}} \). Calculate to find the magnitude of the electric field.

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Key Concepts

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

Energy Density
Energy density is a measure of the amount of energy stored in a particular field per unit volume. It plays a crucial role in both magnetic and electric fields. When we talk about energy density, we're essentially discussing how much energy these fields can "pack" into a specific space.

In magnetic fields, the energy density (\( u_B \)) is calculated using the formula: \[ u_B = \frac{B^2}{2\mu_0}\]where \( B \) is the magnetic field strength and \( \mu_0 \) is the permeability of free space, approximately \( 4\pi \times 10^{-7} \). This formula shows how the energy density increases with the square of the magnetic field strength. Hence, a stronger magnetic field leads to a higher energy density.

In electric fields, the energy density (\( u_E \)) is expressed as:\[u_E = \frac{\varepsilon_0 E^2}{2}\]where \( E \) is the electric field strength, and \( \varepsilon_0 \) is the permittivity of free space, around \( 8.854 \times 10^{-12} \). Similar to magnetic fields, the energy density in an electric field is proportional to the square of the electric field strength.

The key takeaway is that both types of fields store energy, and the way they do so is very similar in structure. For this exercise, understanding energy density helps compare the storage capabilities of both fields.
Magnetic Field Strength
Magnetic field strength is a measure of how intense a magnetic field is in a given space. In the context of our exercise, when the magnetic field strength is given as \( 12 \mathrm{~T} \), it signifies a fairly strong magnetic field, typical in laboratory settings.

The significance of magnetic field strength lies in its contribution to the energy density of the field. As seen in the formula for magnetic energy density:\[ u_B = \frac{B^2}{2\mu_0}\]the field strength \( B \) is squared, meaning any increase in \( B \) dramatically increases the energy stored in the field. This quadratic relationship highlights the efficiency of stronger magnetic fields in storing energy.

Another concept given by magnetic field strength is the notion of magnetic forces and influences in a space. A larger \( B \) denotes a stronger field that can exert more force on charged particles. This principle is crucial in practical applications, where manipulating magnetic fields allows us to control charges and currents effectively.
Electric Field Strength
Electric field strength measures the force felt by a charge within an electric field. It's defined as the electric force per unit charge and is measured in volts per meter \( \mathrm{V/m} \). In this exercise, the goal is to determine what electric field strength yields the same energy density as a given magnetic field.

Electric field strength is integral to calculating energy density in an electric field using the formula:\[u_E = \frac{\varepsilon_0 E^2}{2}\]Again, the dependency on the square of \( E \) means the energy density grows rapidly with increasing field strength.

To solve the exercise, you equate the energy densities: \( u_B = u_E \), leading to:\[\frac{B^2}{2\mu_0} = \frac{\varepsilon_0 E^2}{2}\]By solving this equation, we find that:\[ E = \sqrt{\frac{B^2}{\mu_0 \varepsilon_0}} \]Substituting the known values for \( B \), \( \mu_0 \), and \( \varepsilon_0 \), you'll find the electric field strength required for the same energy density as the magnetic field. This shows the interconnectedness between magnetic and electric fields in terms of energy.

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Most popular questions from this chapter

A circular coil \((950\) turns, radius \(=0.060 \mathrm{~m})\) is rotating in a uniform magnetic field. At \(t=0 \mathrm{~s}\), the normal to the coil is per pendicular to the magnetic field. At \(t=0.010 \mathrm{~s}\) the normal makes an angle of \(\phi=45^{\circ}\) with the field because the coil has made oneeighth of a revolution. An average emf of magnitude \(0.065 \mathrm{~V}\) is induced in the coil. Find the magnitude of the magnetic field at the location of the coil.

In a television set the power needed to operate the picture tube comes from the secondary of a transformer. The primary of the transformer is connected to a wall receptacle. (a) How is the power delivered by the receptacle to the primary related to the power delivered by the secondary to the picture tube? Give your answer in the form of an equation, and explain what assumptions are implied when this equation is used. (b) How is the turns ratio of the transformer related to the currents in the primary and the secondary? (c) How is the turns ratio of the transformer related to the voltage across the primary and the voltage across the secondary? (d) Express the turns ratio \(N_{\mathrm{s}} /\) \(N_{\mathrm{p}}\) of the transformer in terms of the power \(P\) used by the picture tube, the voltage \(V_{\mathrm{p}}\) across the primary, and the current \(I_{\mathrm{s}}\) in the secondary. The primary of the transformer is connected to a \(120-\mathrm{V}\) receptacle. The picture tube of a television set uses \(91 \mathrm{~W}\), and there is \(5.5 \mathrm{~mA}\) of current in the secondary coil of the transformer to which the tube is connected. Find the turns ratio \(N_{\mathrm{s}} / N_{\mathrm{p}}\) of the transformer.

A planar coil of wire has a single turn. The normal to this coil is parallel to a uniform and constant (in time) magnetic field of \(1.7 \mathrm{~T}\). An emf that has a magnitude of \(2.6 \mathrm{~V}\) is induced in this coil because the coil's area \(A\) is shrinking. What is the magnitude of \(\Delta A / \Delta t,\) which is the rate \(\left(\right.\) in \(\left.m^{2} / s\right)\) at which the area changes?

Concept Questions The drawing shows a straight wire carrying a current \(I\). Above the wire is a rectangular loop that contains a resistor \(R\). (a) Does the magnetic field produced by the current \(I\) penetrate the loop and generate a magnetic flux? (b) When is there an induced current in the loop, if the current \(I\) is constant or if it is decreasing in time? (c) When there is an induced magnetic field produced by the loop, does it always have a direction that is opposite to the direction of the magnetic field produced by the current \(I\) ? Provide a reason for each answer. Problem If the current \(I\) is decreasing in time, what is the direction of the induced current through the resistor \(R\) - left to right or right to left? Give your reasoning.

The drawing shows a type of flow meter that can be used to measure the speed of blood in situations when a blood vessel is sufficiently exposed (e.g., during surgery). Blood is conductive enough that it can be treated as a moving conductor. When it flows perpendicularly with respect to a magnetic field, as in the drawing, electrodes can be used to measure the small voltage that develops across the vessel. Suppose the speed of the blood is \(0.30 \mathrm{~m} / \mathrm{s}\) and the diameter of the vessel is \(5.6 \mathrm{~mm} .\) In a 0.60 -T magnetic field what is the magnitude of the voltage that is measured with the electrodes in the drawing?

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