/*! 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 72 Total energy of electromagnetic ... [FREE SOLUTION] | 91影视

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Chapter 16: Problem 72

Total energy of electromagnetic waves in vacuum is given by the relation (A) \(\frac{1}{2} \cdot \frac{E^{2}}{\varepsilon_{0}}+\frac{B^{2}}{2 \mu_{0}}\) (B) \(\frac{1}{2} \varepsilon_{0} E^{2}+\frac{1}{2} \mu_{0} B^{2}\) (C) \(\frac{E^{2}+B^{2}}{c}\) (D) \(\frac{1}{2} \varepsilon_{0} E^{2}+\frac{B^{2}}{2 \mu_{0}}\)

Short Answer

Expert verified
The correct expression for the total energy of electromagnetic waves in a vacuum is (D): \( u = \frac{1}{2} \varepsilon_0 E^2 + \frac{B^2}{2 \mu_0} \).

Step by step solution

01

Identify the expressions for energy density of electric and magnetic fields

In an electromagenetic wave, the energy is stored in both electric and magnetic fields. The energy densities for electric and magnetic fields are given by the relations: Electric field energy density: \( u_{E} = \frac{1}{2} \varepsilon_0 E^2 \) Magnetic field energy density: \( u_{B} = \frac{1}{2} \frac{B^2}{\mu_0} \)
02

Combine the expressions for total energy density

The total energy density in an electromagnetic wave is the sum of the energy densities of the electric and magnetic fields: Total energy density: \( u = u_{E}+u_{B} \) Substitute the expressions for \(u_{E}\) and \(u_{B}\) from Step 1: Total energy density: \( u = \frac{1}{2} \varepsilon_0 E^2 + \frac{1}{2} \frac{B^2}{\mu_0} \)
03

Compare the expression with the given options

By examining the derived expression for total energy density, we can see that our expression matches with option (D): Total energy density: \( u = \frac{1}{2} \varepsilon_0 E^2 + \frac{B^2}{2 \mu_0} \) Therefore, the correct expression for the total energy of electromagnetic waves in a vacuum is (D).

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Most popular questions from this chapter

A small square loop of wire of side \(\ell\) is placed inside a large square loop of wire of side \(L(L \gg \ell)\). The loops are coplanar and their centres coincide. The mutual inductance of the system is proportional to (A) \(\frac{l}{L}\) (B) \(\frac{l^{2}}{L}\) (C) \(\frac{L}{l}\) (D) \(\frac{L^{2}}{l}\)

A conducting \(\operatorname{rod} A C\) of length \(4 \ell\) is rotated about a point \(O\) in a uniform magnetic field directed into the paper. \(A O=\ell\) and \(O C=3 \ell\). Then (A) \(V_{O}-V_{A}=\frac{B \omega l^{2}}{2}\) (B) \(V_{O}-V_{A}=\frac{9}{2} B \omega l^{2}\) (C) \(V_{A}-V_{C}=4 B \omega l^{2}\) (D) \(V_{O}-V_{C}=\frac{9}{2} B \omega l^{2}\)

A magnetic field \(\vec{B}=\left(\frac{B_{0} y}{a}\right) \hat{k}\) is into the paper in the \(+z\) direction. \(B_{0}\) and \(a\) are positive constants. A square loop \(E F G H\) of side \(a\), mass \(m\), and resistance \(R\), in \(x-y\) plane, starts falling under the influence of gravity. Assume \(x\)-axis is horizontal and \(y\) is vertically downward. The magnitude and direction of the net Lorentz force, acting on the loop when its speed is \(v\), is (A) \(\frac{B_{0} a^{2} v}{R}\), upward (B) \(\frac{B_{0} a^{2} v}{R}\), downward (C) \(\frac{B_{0}^{2} a^{2} v}{R}\), downward (D) \(\frac{B_{0}^{2} a^{2} v}{R}\), upward

A conducting rod \(A B\) moves parallel to \(x\)-axis in the \(x-y\) plane. A uniform magnetic field \(B\) pointing normally out of the plane exists throughout the region. A force \(F\) acts perpendicular to the rod, so that the rod moves with uniform velocity \(v\). The force \(F\) is given by (neglect resistance of all the wires). (A) \(\frac{v B^{2} l^{2}}{R} e^{-t / R C}\) (B) \(\frac{v B^{2} l^{2}}{R}\) (C) \(\frac{v B^{2} l^{2}}{R}\left(1-e^{-t / R C}\right)\) (D) \(\frac{v B^{2} l^{2}}{R}\left(1-e^{-2 t / R C}\right)\)

When the current changes from \(+2 \mathrm{~A}\) to \(-2 \mathrm{~A}\) in \(0.05\) second, an EMF of \(8 \mathrm{~V}\) is induced in a coil. The coefficient of self-induction of the coil is (A) \(0.2 \mathrm{H}\) (B) \(0.4 \mathrm{H}\) (C) \(0.8 \mathrm{H}\) (D) \(0.1 \mathrm{H}\)
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