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All Solutions of Type: Electromagnetism
 0 Click here to jump to the problem! GR8677 #3 Electromagnetism$\Rightarrow$}Wave Equation Simply and elegantly stated: $\frac{1}{c^2}\frac{\partial^2 \phi}{\partial t^2}=\frac{\partial^2 \phi}{\partial x^2}$. (One can check this by noting that the dimensions cancel out to $meters^{-2}$ on both sides.) Now, $c=\frac{1}{\sqrt{\epsilon_0 \mu_0}}$. Elsewhere than a vacuum, it's $v=\frac{1}{\sqrt{\epsilon \mu}}$, where, in this problem, we have $\epsilon=2.1\epsilon_0$ and $\mu=\mu_0$. \par No need to narrow the choices down. (D) is right because $v=\frac{1}{\sqrt{2.1 \times 1 \epsilon_0 \mu_0}}=\frac{c}{\sqrt{2.1}}$, where the last equality comes from substituting the definition of the speed of light (via an epsilon and a mu). Click here to jump to the problem!
 1 Click here to jump to the problem! GR8677 #9 Electromagnetism$\Rightarrow$}Current Density There are (at least) two ways to solve this problem. If one has a decent memory or might make for a decent experimenter, one should remember that the electron drift velocity is on the order of $10^{-4}m/s$. Choice D is the only thing that fits this order. If one is a theorist who can't remember jack, then one would have to do a bit more work: Recall the basic equation: $j=\frac{I}{A}=nev,$ where $n=\frac{N}{V}=1E28$, i.e., the number of electron per volume, and $v$ is the drift velocity. The area is $A=\pi(\frac{d}{2})^2$. Solving for $v$, and making the appropriate approximate plug-in's: $\frac{4I}{\pi d^{2}}\frac{1}{ne}\approx\frac{4I}{\pi (\frac{2}{100})^{2}}\frac{1}{(1E28)(1.5E-19)}\approx 3E-4, $ which is closest to choice D. Click here to jump to the problem!
 2 Click here to jump to the problem! GR8677 #10 Electromagnetism$\Rightarrow$}Gauss' Law Only choices C, D, and E make sense, since the field at $r=0$ has to be 0. Moreover, since a uniform charge distribution is applied, the field has to increases linearly within the volume of the sphere. Choice D and E are both out, leaving just C. Direct verification via applying Gauss' Law for the inside of the sphere: $E(4\pi r^2)=Q=\rho \frac{4}{3}\pi r^3\Rightarrow E=\frac{\rho r}{3}\propto k r$. Outside, all five choices converge to an inverse-square decreasing field---so, no worry there! Click here to jump to the problem!
 3 Click here to jump to the problem! GR8677 #11 Electromagnetism$\Rightarrow$}Vector Calculus There are two identities from vector calculus one has to know by heart. The one directly applicable to this problem is: $\nabla \cdot (\nabla \times \vec{H})=0$ Plug in the equation given in the problem to the identity above to get 0. (The other identity, not quite as useful for this problem, but perhaps useful for subsequent problems, is: $\nabla\times(\nabla f)$) Click here to jump to the problem!
 4 Click here to jump to the problem! GR8677 #22 Electromagnetism$\Rightarrow$}Lorentz Transformation When an electric field is Lorentz-transformed, afterwards, there might be both a magnetic and electric field (in transverse components). Or, more rigorously, one has, $\begin{eqnarray} E_x'&=& E_x\\ E_y'&=& \gamma(E_y-vB_z)\\ E_z'&=& \gamma(E_z+vB_y)\\ B_x' &=& B_x\\ B_y' &=& \gamma(B_y + E_z v/c^2)\\ B_z' &=& \gamma(B_z-vE_y/c^2), \end{eqnarray}$ for motion in the $x$ direction. (A) Obviously not. Suppose initially, one has just $E_x$, afterwards, there's still $E_x'$. (B) True, as can be seen from the equations above. (C) Not true in general. Suppose $\vec{E}=E_x\hat{x}+E_y\hat{y}+E_z\hat{z}$. Afterwards, the transverse components are off by a $\gamma$ even if there's no B field to start with. (D) Nope. (E) Mmmm... no need for gauge transformations. Click here to jump to the problem!
 5 Click here to jump to the problem! GR8677 #24 Electromagnetism$\Rightarrow$}Circuits Apply Kirchhoff's Loop Law: $(G-V)-I(R+1)=0$, where $G$ is the voltage of the generator and $V$ is the voltage of the battery, and the extra $1$ is the internal resistance. $I=10Amperes$ in the problem, and thus $R=1\Omega$. Click here to jump to the problem!
 6 Click here to jump to the problem! GR8677 #25 Electromagnetism$\Rightarrow$}Lorentz Force Recall the Lorentz Force, $\vec{F}=q(\vec{E}+\vec{v}\times\vec{B})$. $\vec{E}$ and $\vec{B}$ are parallel. The particle is released from rest, so the Electric force would propel it. The resulting velocity would be parallel to the electric field, but since the magnetic field is also parallel to that, there would be no magnetic force contribution. The particle thus goes in a straight line. Click here to jump to the problem!
 7 Click here to jump to the problem! GR8677 #29 Electromagnetism$\Rightarrow$}Right Hand Rule Note: this is a negative test charge. So, for the test charge, one has to either choose the opposite direction as that yielded by the Right Hand Rule or one could use the Left Hand Rule, which is just the RHR done with the left hand. Suppose the current is a straight line pointing upwards along the page. The RHR for the current shows a magnetic field that's coming out of the page from the left side of the current and a field going into the page on the right side. The problem wants the test charge to go parallel to the current. Applying the Lorentz Force, where $F\propto \vec{v}\times \vec{B}$, one finds that no matter the direction of approach, the only way for the force to point parallel to the current is for the velocity to go towards the wire. (Check this: Suppose the charge comes in from the left; the force would point parallel to the current. Suppose the charge comes in from the right; the force would point parallel to the current, again.) Click here to jump to the problem!
 8 Click here to jump to the problem! GR8677 #52 Electromagnetism$\Rightarrow$}Potential By Gauss Law, if there are no charges inside the cube, then the electric field inside would be 0. The potential $\phi$ is related to the field $\vec{E}$ by $\vec{E} = \nabla \phi$, and thus since $\vec{E}=0$, one infers that $\phi$ is constant. Since the potential function has to remain continuous its value everywhere inside of the cube is the same as that at the surface of the cube, which is given as $V$. Click here to jump to the problem!
 9 Click here to jump to the problem! GR8677 #53 Electromagnetism$\Rightarrow$}Accelerating Charges One doesn't really need to understand the mess Griffiths has on accelerating charges to solve this problem. In fact, common sense works. Since the particle spends the most time about the origin, the field there should be maximal. The origin corresponds to $\theta=90^\circ$, via the system used in the diagram. This is choice (C). Click here to jump to the problem!
 10 Click here to jump to the problem! GR8677 #54 Electromagnetism$\Rightarrow$}Dielectric Recall the following equations, $\nabla \cdot \vec{D} = \rho_f /; /; /; \vec{D}=\epsilon_0 \vec{E} + \vec{P}=\kappa \epsilon_0 \vec{E}.$ Also, recall the relation between bound charge and polarization, $\nabla \cdot \vec{P} = -\sigma_b.$ Use the divergence theorem on the above equations to apply the elementary Gauss' Law to the region, $\oint \vec{D} \cdot dA = \int \sigma dA \Rightarrow D=\sigma. $ $\oint \vec{P} \cdot dA = -\int \sigma_p dA \Rightarrow P=-\sigma_p. $ But, since $\vec{D}=\epsilon_0 \vec{E} + \vec{P}=\kappa \epsilon_0 \vec{E}$, one has $D=\sigma=\kappa \epsilon_0 E\Rightarrow E=\frac{\sigma}{\kappa \epsilon_0}$. Plug that into the other relation for D (and use the result for P from above) to get, $\sigma = \frac{\sigma}{\kappa} - \sigma_p$. Thus, $\sigma_p = \sigma \frac{1 - \kappa}{\kappa}$, as in choice (E). Click here to jump to the problem!
 11 Click here to jump to the problem! GR8677 #63 Electromagnetism$\Rightarrow$}Method of Image Charge The boundary condition at the conducting plane is $V(0)=0$. This doesn't mean that one can't put an image charge" a distance $-D$ away on the other side of the plane to make the calculation easier. Making the directest straight line from the charge to the plane along the $z$ axis, one gets the following image-charge potential: $V = \frac{1}{4\pi\epsilon_0} \left(\frac{q}{\sqrt{x^2+y^2+(z-d)^2}}-\frac{q}{\sqrt{x^2+y^2+(z+d)^2}}\right).$ The charge density for a grounded conducting plane is given by $\rho=-\epsilon_0 \frac{\partial V}{\partial n}$, where $n$ is the direction of the directest straight line. The derivative is, $\frac{\partial V}{\partial z}=\frac{1}{4\pi\epsilon_0} \left(\frac{q(z-d)}{\left(x^2+y^2+(z-d)^2\right)^1.5}-\frac{q(z+d)}{\left(x^2+y^2+(z+d)^2\right)^{1.5}}\right),$ and thus, after combining terms, the density is $-\frac{1}{2\pi}\frac{qd}{\left(x^2+y^2+(z-d)^2\right)^{1.5}}$ The problem wants the surface charge density a distance $D$ away from the point charge. So, plug in $\left(x^2+y^2+(z-d)^2\right)^{1.5}=(D^2)^{1.5}$ to get $-\frac{1}{2\pi}\frac{qd}{D^3}$ Click here to jump to the problem!
 12 Click here to jump to the problem! GR8677 #64 Electromagnetism$\Rightarrow$}Impedance Matching Because the capacitor is fully charged, it is the generator. The inductor, which opposes the current as the capacitor discharges, is the load. Impedance matching requires that the impedance of the generator and load are the same. However, since $Z_g=R_g+jX_g$ and $Z_L = R_l + j X_l$, for the generator and load, respectively, and $X_g = X_{gL} - X_{gC} = -X{gC}$ and $X_l = X_{lL} - X_{lC} = X_{IL}$, one finds that for impedance matching, $X_g = -X_l$. More info here Click here to jump to the problem!
 13 Click here to jump to the problem! GR8677 #65 Electromagnetism$\Rightarrow$}Biot-Savart Law One either remembers the field of a circular current loop or one derives it from the Bigot-Savart Law, $\vec{B} = \frac{\mu_0}{4\pi}\frac{I d\vec{l}\times\hat{r}}{r^2}.$ One finds that $d\vec{l} \times \hat{r} = dl$, as they're perpendicular. $dB = \frac{\mu_0 I dl}{4\pi r^2}$, and the vertical components of the field cancel, thus there are only the horizontal components (parallel to the axis of the ring). $r$ is the directest distance from a differential element on the wire to a point on the axis, and $x$ is the horizontal distance (along the axis) from the center of the ring to that point. Thus, the integral becomes, $\int \frac{\mu_0 I dl}{4\pi}\frac{b}{(x^2+b^2)^{3/2}} = \frac{\mu_0 I b}{2 }\frac{b}{(x^2+b^2)^{3/2}}, $ where $\int dl=2\pi b$. So anyway, the field, at a point anywhere along the axis of the loop is $\vec{B}=\frac{\mu_0 I b}{2 }\frac{b}{(x^2+b^2)^{3/2}}$. At a point far, far, away, $x >> b$, and thus $\vec{B}\approx \frac{\mu_0 I b}{2 }\frac{b}{x^3} = \frac{\mu_0 I b^2}{2 x^3}$. Set $x=c$, where $c$ is the fixed coordinate of that point, to get that the field is proportional to $Ib^2$. Incidentally, defining the magnetic dipole moment to be $\mu=IA=I\pi b^2$, one finds the field of a magnetic dipole to be $\frac{\mu \mu_0}{2\pi x^3}$. The field far from a current loop is the same as the field of a magnetic dipole. Click here to jump to the problem!
 14 Click here to jump to the problem! GR8677 #79 Electromagnetism$\Rightarrow$}Field Lines The Maxwell Equation that states No monopoles" requires that the divergence of the magnetic field be 0, or more elegantly, $\nabla \cdot \vec{B} = 0$. The problem asks for fields that violate this condition, so the condition to look for now is $\nabla \cdot \vec{B} \neq 0$. (A) $\vec{B}=\pm Const$ ... divergence is 0 (B) $\vec{B}=\pm Const$ ... divergence is 0 (C) this does not explicitly require the divergence to be 0. (D) this is a non-zero divergence, as the field lines diverge outwards from a source. (E) this is $\nabla \times \vec{B}\neq 0$... doesn't necessarily state anything about $\nabla \cdot \vec{B}$ Click here to jump to the problem!
 15 Click here to jump to the problem! GR8677 #80 Electromagnetism$\Rightarrow$}Gauss Law Recall the differential form of Gauss' Law $\nabla \cdot \vec{E}=\rho$. For fields that contain no charges, the equation becomes $\nabla \cdot \vec{E}=0$. Find the choice with 0 divergence. (A) $A(2y-x) \neq 0$ (B) $A(-x+x)=0$ ... so this is it! (C) $A(z) \neq 0$ (D) $A(yz+xz) \neq 0$ (E) $Ayz \neq 0$ Click here to jump to the problem!
 16 Click here to jump to the problem! GR8677 #81 Electromagnetism$\Rightarrow$}Faraday's Law Recall Faraday's Law, $V=-\frac{d\Phi}{dt},$ where $\Phi=\vec{B}\cdot d\vec{A}$. The magnetic field at the center of a loop of current-carrying wire is $\vec{B}=\frac{\mu_0 I}{2 r}$, where $r$ is the radius of the loop. (If one forgot this, one can remember it from Ampere's Law.) The magnetic field of the outer loop induces an electric field in the inner loop. The outer loop's B field is $\vec{B}=\frac{\mu_0 I}{2 b}$. The area of the inner loop stays constant at $\pi a^2$. Thus the flux through it is $\Phi = \frac{\mu_0 I}{2 b} \pi a^2$. The larger loop carries an ac current, given by $I=I_0 \cos\omega t$. Thus, $|\frac{d\Phi}{dt}|=\frac{\mu_0 I_0 \omega}{2 b} \pi a^2 \sin\omega t,$ as in choice (B). Click here to jump to the problem!
 17 Click here to jump to the problem! GR8677 #86 Electromagnetism$\Rightarrow$}Coulomb's Law (Better classified as the Cavendish-Maxwell Experiment to determine the exponent in Coulomb's Law.) One wonders why the common sense (and much too trivial) answer, choice (D), isn't right. In searching through one's lower level textbook, one would find accounts of Cavendish's Torsion experiment, which seems to also support choice (D). However, reading up on papers, one finds that the precise determination of the exponent is actually done via a whole different method... which although is an experimental technique, actually serves to illuminate the necessity of the inverse square law---and thus, this problem is not classified as a Lab Technique problem. Charge up a conducting shell. Put a charge inside, distinctly asymmetrically far from the center but not touching the outer shell. One finds that the charge will not move. This suggests that the field is 0. Suppose that the field contribution to the test charge can be broken down into two unequal parts, set apart by a plane through the test-charge. $\vec{E}\propto dS_1/r_1^n - dS_2/r_2^n$. Substituting the solid angle $d\Omega = dS \cos\theta/r^2$, one finds that $\vec{E}\propto \frac{d\Omega}{\cos\theta}\left( \frac{1}{r_1^{n-2}} - \frac{1}{r_2^{n-2}}\right)$. Since the field vanishes for any $r_1$, $r_2$, one deduces that $n\rightarrow 2$. For other and more modern methods, see page 95ff, especially page 100 of $http://yosunism.com/inverseSquare.pdf$ Click here to jump to the problem!
 18 Click here to jump to the problem! GR8677 #94 Electromagnetism$\Rightarrow$}LR Circuit Once the switch S is closed, although the initial current through the inductor is 0, the change in current through it is maximal. Recalling that $V = -L \dot{I}$, one realizes that the voltage across $A$ must be non-zero at the start. Thus, plots (C), (D), and (E) are eliminated. Now, one must decide between choices (A) and (B). Once the circuit reaches equilibrium, i.e., the elements reach their asymptotic values. Specifically, the voltage at A goes to 0, since the inductor has no potential difference across it (to wit $\dot{I}=0$). Once the switch is opened, the current suddenly changes, and $\dot{I} \neq 0$, thus the inductor has a voltage across it, and the voltage at A becomes nonzero. Because of the diode, this would necessarily have to be a negative voltage. Since there is only $R_2$ to dissipate the voltage, the magnitude of the voltage once the switch is opened should be bigger than that initially, when the switch is closed at $t_0$. Choose (B). Click here to jump to the problem!
 19 Click here to jump to the problem! GR8677 #98 Electromagnetism$\Rightarrow$}Potential Recall that $V=\int \frac{dq}{x}=k\int_l^{2l} \lambda \frac{dx}{x}=k\lambda\ln 2$, where $\lambda=Q/l$ is your usual linear charge density. Click here to jump to the problem!
 20 Click here to jump to the problem! GR8677 #1 Electromagnetism$\Rightarrow$}RC Circuit One can immediately eliminate plots A, C, E, since one would expect an exponential decay behavior for current once the switch is flipped. More rigorously, the initial circuit with the switch $S$ connected to $a$ has the following equation, $V-\dot{Q} r - Q/C =0 \Rightarrow \frac{dQ}{dt}=\frac{1}{r}\left(V-Q/C\right)$ Once integrated, the equation becomes, $Ve^{-\frac{t}{rC}} = V-Q/C \Rightarrow Q/C = V\left(1-e^{-\frac{t}{rC}} \right)$ The charge stored on the capacitor after it is fully charged is $Q=CV$ (at $t=0$). When the switch $S$ is switched to $b$, the equation becomes, $Q/C = \dot{Q}R \Rightarrow Q(t) = Q_0 e^{-\frac{t}{RC}},$ where $Q_0=CV$ from the initial connection. Current is the negative time derivative of charge, and thus, $I(t)=-\dot{Q}=\frac{Q_0}{RC} Q(t) = \frac{V}{R}Q(t)$ The initial current IS $I(0)=V/R$, and thus choice (B) is right. Click here to jump to the problem!
 21 Click here to jump to the problem! GR8677 #2 Electromagnetism$\Rightarrow$}Faraday Law Recall Faraday's Law $\nabla \times \vec{E} = -\frac{\partial \vec{B}}{\partial t}$. Dot both sides with the area $d\vec{A}$. Recalling Stokes' Theorem ($\int \nabla \times \vec{E} \cdot d\vec{A}=-\frac{\partial \vec{B}}{\partial t}\cdot d\vec{A}$), the left side can be converted to the potential, i.e., the emf $\mathcal{E}=-\int \vec{E} \cdot d\vec{l}=-\int \nabla \times \vec{E} \cdot d\vec{A}$. Finally, from Ohm's Law $V-\mathcal{E}=IR$, one can obtain the current. (Note that $V=5.0$ V is the voltage of the battery. The voltage induced acts to oppose this emf from the battery.) The problem gives $\frac{dB}{dt}=150T/s$. The area is just $0.1^2m^2$. Thus, the induced emf is, $\mathcal{E}= \frac{d B}{d t} A = 150/100 =1.5$ Thus, $V-\mathcal{E}=3.5=IR \Rightarrow I=0.35A$, since $R=10 \Omega$. Click here to jump to the problem!
 22 Click here to jump to the problem! GR8677 #3 Electromagnetism$\Rightarrow$}Potential Recall the elementary equations, $V = \int \vec{E}\cdot d\vec{l} = \int \frac{dq}{4\pi \epsilon_0 r}$. $r=\sqrt{R^2+x^2}$, and $dQ=Q$ thus $V=\frac{Q}{4\pi \epsilon_0 \sqrt{R^2+x^2}}$, as in choice (B). Click here to jump to the problem!
 23 Click here to jump to the problem! GR8677 #4 Electromagnetism$\Rightarrow$}Small Oscillations The potential is determined in the previous problem to be $V=\frac{Q}{4\pi \epsilon_0 \sqrt{R^2+x^2}}$. The field is given by $\vec{E}=-\nabla V$. Before taking derivatives, one can simplify the potential since it is given that $R >> x$. Binomial expand it ($(1+y)^n\approx 1+ny$, for $y$ small) to get $V\approx \frac{Q}{4\pi \epsilon_0 R}(1+\frac{x^2}{2R^2})$ Taking the derivative, using the equations $\vec{E}=-\nabla V$, and $\vec{F}=q\vec{E}$, one gets, $-q\frac{Q}{4\pi \epsilon_0 R}\frac{x}{R^2}=F=m\ddot{x}.$ Small oscillations have the same form as simple harmonic oscillations, i.e., $\ddot{x} = -\omega^2 x$. The angular frequency is $\omega=\sqrt{q\frac{Q}{4\pi \epsilon_0 R^3 m}}$, as in choice (A). Click here to jump to the problem!
 24 Click here to jump to the problem! GR8677 #19 Electromagnetism$\Rightarrow$}Coulomb's Law The particle obeys a Coulomb's Law potential, $V=\frac{k Z_1 Z_2 q^2}{r}$. In this case particle 1 is a Helium atom, which has charge $Z_1=2$, while particle 2 is silver, with $Z_2=50$. Thus, $V=\frac{100 k q^2}{r}.$ Conservation of energy requires that when the incident particle is at its closest approach, $5MeV=\frac{100k q^2}{r}$. Recall that $k=9E9$, $q=1.602E-19$, convert everything to $SI$ to get $r\approx 2.9E-14$. Click here to jump to the problem!
 25 Click here to jump to the problem! GR8677 #24 Electromagnetism$\Rightarrow$}Conductors This problem involves applying Coulomb's Law $F\propto q_A q_B/r_{AB}^2$ to conductors. The charge travels from conductor to conductor and equilibriates instantaneously due to the requirement that two touching conductors must be at an equipotential. This means that if conductors 1 and 2 touch then their potentials are related by $V_1=V_2$. Because the problem involves spherical conductors, the potential has the form $V\propto q_A q_B/r_{AB}$. The initial force between the two conductors is $F$, where $q_A=q_B=Q$. After C is touched to A, the charge becomes $q_A=Q/2=q_C$, since each conductor shares the same charge out of a total of $Q$ (to wit: each has half of the total charge). When C is touched to B, the charge becomes $q_C=3/4Q=q_B$, since each conductor shares the same charge out of a total of $Q+Q/2$ (to wit: $\frac{1}{2}3/2 Q = 3/4 Q$ for each conductor). When C is removed, one calculates the force from Coulomb's law and the final charges on A and B determined above to be, $F = 3/8 Q^2/r_{AB}^2=3/8 F$, as in choice (D). Click here to jump to the problem!
 26 Click here to jump to the problem! GR8677 #25 Electromagnetism$\Rightarrow$}Capacitors Recall the following truths (held to be self-evident?) on the subject of capacitors: 1. series capacitors have equal charge (Equivalent capacitance of two capacitors is $1/C_{eq} = 1/C_1 + 1/C_2$); 2. parallel capacitance have equal voltage ($C_{eq}=C_1+C_2$); 3. $Q=CV$; 4. $U=\frac{1}{2}C V^2$. (A) Initially, before the switch is closed, only $C_1$ has a voltage across it, and hence it is charged. $Q_0=CV$. But, afterwards, since the voltage stays the same, one has $Q_1=Q_2=CV$; hence, $Q_0=\frac{1}{2}(Q_1+Q_2)$. (B) $V_1=V_2\Rightarrow Q_1/C_1 = Q_2/C_2$. Since $C_1=C_2$, one has $Q_1=Q_2$. This is true. (C) By definition of circuit elements in parallel, one has each capacitor at the same potential. This is trivially true. $V_1=V_2=V$ (D) Since one determined from (C) that the capacitors are at the same voltage, then because they have the same capacitance, they have the same energy as per $U=\frac{1}{2}C V^2$. True. (E) This is false, since $U_0=\frac{1}{2}CV^2$, initially. In the final state, each capacitor has energy $\frac{1}{2}CV^2$. The sum of energies is thus $2U_0$. Click here to jump to the problem!
 27 Click here to jump to the problem! GR8677 #26 Electromagnetism$\Rightarrow$}Resonance Frequency One wants to tune one's radio to the resonance frequency (a.k.a. the frequency at which impedance is matched). The resonance frequency of an LRC circuit is given by $\omega^2 = 1/LC$, where the quantities involved are angular frequency, inductance, and capacitance. Solving for C, one has $C=1/(L\omega^2) \approx 1/(2E-6 * 36 * 100E6) =1/(7.2E-11)\approx 0.1E-11 =1E-12$. This is choice (C). The hardest part of his problem, of course, is doing the math without a calculator. Easy. Click here to jump to the problem!
 28 Click here to jump to the problem! GR8677 #43 Electromagnetism$\Rightarrow$}Stokes Theorem Recall Stokes' Theorem $\oint \nabla \times u d\vec{a} = \int \vec{u} \cdot d\vec{l}$. The left side of the equality is easier to evaluate, so evaluating that, one has $\nabla \times u =2$. The area is $\pi R^2$, and thus $\int \vec{u} \cdot d\vec{l}= 2\pi R^2$. Click here to jump to the problem!
 29 Click here to jump to the problem! GR8677 #46 Electromagnetism$\Rightarrow$}Faraday Law Recall Faraday Law, $\epsilon = -\frac{d\Phi}{dt}$, where $\Phi=B\cdot dA$. Since the magnetic field is constant, the equation simplifies to $\epsilon = -B \cdot \frac{dA}{dt}$ for this case. $B \cdot A = B \cos(\omega t) \pi r^2$, and thus $d\Phi/dt = -\omega \sin(\omega t) \pi r^2 = -\epsilon = -\epsilon_0 \sin(\omega t)$. Solving for angular momentum, one has $\omega = \epsilon_0/(B\pi R^2)$. Alternatively, one has $A=\pi r r(t) \Rightarrow \dot{A} = \pi r \dot{r} = \pi r v$. Since $v=\omega r\sin(\omega t)$, one has $A=\pi r \omega r\sin(\omega t)$. Plug it into Faraday Law and solve for angular velocity. Click here to jump to the problem!
 30 Click here to jump to the problem! GR8677 #47 Electromagnetism$\Rightarrow$}Faraday Law Recall Faraday's Law, $\epsilon=\vec{E} \cdot d\vec{l}=-d\Phi/dt$, where $\Phi = \vec{B} \cdot d\vec{A}$. In words, this means that a changing flux (either a varying field or radius) induces a voltage. The field is given as just B. The area of the loop is just $\pi R^2$, i.e., the cross-sectional area of the cylinder. As the cylinder is spun around, its flux changes at the rate of N rps. The change in flux is thus $N B \pi R^2$, and this is the magnitude of the potential difference in choice (C). (Also, one can drop out the other choices from units. And, since the cylinder is moving in a magnetic field, the non-zero flux demands a voltage, so (A) can't be it.) Click here to jump to the problem!
 31 Click here to jump to the problem! GR8677 #49 Electromagnetism$\Rightarrow$}Relativistic Fields For motion in the x direction, one has the following equations for the E and B fields, $\begin{eqnarray} E_x &=& E_x'\\ B_x &=& B_x'\\ E_y &=& \gamma (E_y' - v B_z')\\ B_y &=& \gamma (B_y' - v/c^2 E_z')\\ E_z &=& \gamma (E_z' - v B_y')\\ B_z &=& \gamma (B_z' - v/c^2 E_y') \end{eqnarray}$ Since $E_z' = \sigma/(2\epsilon_0)$ (with all other primed components 0), the transformed field is just $E_z=\gamma E_z'$, as in choice (C). (Recall that $\gamma = 1/\sqrt{1-\beta^2}$, where $\beta = v/c$) If one forgets the Lorentz-transformed fields, one can also quickly derive the answer for this case. Since the transformed charge density is Lorentz contracted in one of its area dimensions, one has $\sigma = \gamma \sigma '$. One can tell by symmetry of the surface that the other field components cancel, and one again arrives at the result for $E_z$ as above. Click here to jump to the problem!
 32 Click here to jump to the problem! GR8677 #61 Electromagnetism$\Rightarrow$}Gauss Law Recall Gauss Law $\vec{E} \cdot d\vec{A} = q/\epsilon_0$. Thus, $E(4\pi r^2)=\int_0^{R/2}A r^2 4\pi r^2 dr = 4\pi A (R/2)^5/5$. Solving for E, one has $E=A/5(R/2)^3$, as in choice (B). Click here to jump to the problem!
 33 Click here to jump to the problem! GR8677 #62 Electromagnetism$\Rightarrow$}Capacitors Initially, one has the two capacitors connected in parallel, so that each receives the same voltage from the battery. Thus $V=5=Q_1/C_1=Q_2/C_2$. $C_1=1$ and $C_2=2$, thus $Q_1=C_1 V = 5$ and $Q_2=C_2 V = 10$. After the battery is removed and the capacitors are re-connected so that the opposite plates face each other, one has (immediately) $Q_1 - Q_2 = -5$. The charges would then redistribute themselves so that the voltage across each capacitor is the same. Thus, denoting the final charge on each capacitor as $q_1,q_2$, respectively, one has (from charge conservation) $-5=q_1+q_2$. Applying the equi-voltage condition, one has $q_1/C_1 = q_2/C_2 \Rightarrow q_1 = q_2 C_1/C_2$. Plug that into the charge conservation equation to get $-5=(1+C_1/C_2)q_2\Rightarrow q_2=3.33 \Rightarrow V=q_2/C_2 \approx 1.7$, as in choice (C). (As an exercise, one can also check by computing the charge for the other capacitor.) Click here to jump to the problem!
 34 Click here to jump to the problem! GR8677 #69 Electromagnetism$\Rightarrow$}Ampere Law Recall Ampere's Law, $\vec{B} \cdot d\vec{l} = \mu_0 I_{in$. Since the region one is interested in is a vacuum, one's Ampere Loop encloses all of the current. Thus, the field from each conductor is $B(2\pi r) = \mu_0 J \pi R^2$, where, $I_{in = J \pi R^2$ and R is the radius of the conductor. (This is a good approximation of the current, as one assumes that the vacuum region in the center is small compared to the area of the conductors.) Making the approximation that $R\approx d/2$, one has $B = \frac{\mu_0 J \pi R^2}{2\pi R}\approx \frac{\mu_0 J \pi d}{4\pi}$. Since both fields contribute in the center, the field is twice that, $\frac{\mu_0 J \pi d}{2\pi}$, as in choice (A). (Also, one can immediately eliminate all but choices (A) and (B) by the right-hand rule. One seeks a +y-direction field.) Click here to jump to the problem!
 35 Click here to jump to the problem! GR8677 #70 Electromagnetism$\Rightarrow$}Larmor Formula The Larmor formula for power radiated by an accelerated charge is related to the charge and acceleration as $P \propto a^2 q^2$. The problem gives the following: A: $q,b,a \Rightarrow P_A\propto q^2a^2$ B: $2q,3v,4a \Rightarrow P_B\propto (2q)^2(4a)^2=64q^2a^2$ Thus, $P_B/P_A=64$, as in choice (D). Click here to jump to the problem!
 36 Click here to jump to the problem! GR8677 #71 Electromagnetism$\Rightarrow$}Particle Trajectory One can get a reasonable approximation for the deflection angle as follows. Assuming that there is no magnetic field, one has from the Lorentz force $F=ma=qE=qV/d$, where one neglects gravitational acceleration. The acceleration is constant, and it is $a=qV/(dm)$. Recalling the baby physics kinematics equation, $y=0.5at^2 \Rightarrow dy=atdt$ and the fact that $x=L=vt \Rightarrow dx=vdt$ and $t=L/v$, one can calculate the angle as $\tan \theta \approx dy/dx = \frac{atdt}{vdt}=at/v=\frac{qVL}{v^2dm}$. Take the arctangent to get choice (A). Click here to jump to the problem!
 37 Click here to jump to the problem! GR8677 #86 Electromagnetism$\Rightarrow$}Particle Trajectory There is a force pointing upwards from the Electric field in the y-direction. Suppose the particle is initially moving upwards. Then, the magnetic field would deflect it towards the right... One can apply the Lorentz Force to solve this problem. If the particle comes in from the left, then the magnetic force would initially deflect it downwards, while the electric force would always force it upwards. Continue applying this analysis to each diagram. It turns out that one has cycloid motion whenever the electric and magnetic fields are perpendicular. Click here to jump to the problem!
 38 Click here to jump to the problem! GR8677 #87 Electromagnetism$\Rightarrow$}Faraday Law From Faraday's Law or Len's Law, one has $\vec{E}\cdot d\vec{l} = -d\Phi/dt$. Since, in order for the balls to move, they must move in a circle, one has $dl=2\pi d/2$; moreover, the induced magnetic field would point in the opposite direction to the field that was before, and one has a current in a loop from the right-hand-rule. The area of the magnetic flux is just $\pi R^2$, since the field only goes through the cylindrical region of radius R. Thus, $E(2\pi d/2 ) = \dot{B}\pi R^2 \Rightarrow E=\frac{\dot{B}R^2}{d}$. Now, recall some mechanics. The torque is related to the moment-arm and force by $\tau = \sum \vec{r} \times \vec{F}$, where $\vec{F}=q\vec{E}=q\frac{\dot{B}R^2}{d}$. Since there is a force contribution from each charge, and since, by the right-hand-rule, their cross-products with the moment-arm point in the same direction, one finds the torque to be $\tau=2(d/2)q\frac{\dot{B}R^2}{d} = dq\frac{\dot{B}R^2}{d}=q\dot{B}R^2$. Now, recall the relation between angular momentum and torque to be $\sum \tau = \dot{L}$. Replace the $\dot{B}\rightarrow B$ above to get $L = qBR^2$, and so the system starts rotating with angular momentum as in choice (A). (This approach is due to Matt Krems.) Note that one can immediately eliminate choice (D) since angular momentum is not conserved from the external torque induced (to wit: electromagnetic induction). Moreover, although choice (E) is true in general, it does not apply to this problem. Click here to jump to the problem!
 39 Click here to jump to the problem! GR8677 #88 Electromagnetism$\Rightarrow$}Ampere Law Recall Ampere's Law $\vec{B} \cdot d\vec{l} = \mu_0 I_{in}$, where $I_{in}$ is the current enclosed by the loop dl. Apply it to the region between a and b, $B(2\pi r) = \mu_0 I \frac{\pi r^2 l}{\pi R^2 l} \Rightarrow B = \mu_0 I \frac{r}{2 R^2 \pi}$, which gives a linearly increasing field, and thus choices (D) and (E) and (A) are eliminated. Choices (B) and (C) remain. Apply Ampere's Law to the region outside of the outer sheath. For $r>c$, one has $I_{in}=0 \Rightarrow B(2\pi r) = 0 \Rightarrow B=0$. Choice (B) shows the behavior of zero-field outside the sheathed coax cable. Choose that. Click here to jump to the problem!
 40 Click here to jump to the problem! GR8677 #89 Electromagnetism$\Rightarrow$}Trajectory The only physics involved in this problem is equating the centripetal force with the Lorentz Force, $mv^2/R=qvB$. The rest is math manipulation and throwing out terms of ignorable order. The radius of curvature used in the centripetal force equation is given by $R^2=l^2+(R-s)^2$, and ETS is nice enough to make this geometry fairly obvious in the diagram enclosed with the original question. Now, note that since $s<, after expanding the expression for $r^2$, one can drop out terms of higher order. Thus, $R^2=l^2+(R-s)^2=l^2+R^2+s^2-2Rs\approx l^2+R^2-2Rs + O(s^2)$. Canceling the R's on both side, one finds, $l^2=2Rs \Rightarrow R = l^2/(2s)$. Plug this into the force equation above to find, $mv/R = qB \Rightarrow 2smv/l^2=qB \Rightarrow p=mv=qbl^2/2s, $ which is choice (D). Click here to jump to the problem!
 41 Click here to jump to the problem! GR8677 #7 Electromagnetism$\Rightarrow$}Field Lines Since one has same polarity magnet ends at the same side, one can immediately cancel out choices (A) and (D). They would repel each other, and thus the field-lines would not be pointing towards each other. Since there are no such things as magnetic monopoles, one cannot have choice (C), as that would imply a non-zero divergence in the magnetic field. Choice (D) would work if there were a current through the magnets, but there's not. Choice (B) remains. It makes sense since the magnetic charges should repel each other, and thus the field lines should point away showing the repulsion. Click here to jump to the problem!
 42 Click here to jump to the problem! GR8677 #8 Electromagnetism$\Rightarrow$}Conductors A conductor has the magical mystical property of inducing the exact opposite charge of a nearby charge. One has all the electrons in the conductor moving towards the positive charge placed outside Q. A net negative charge of -Q is induced. (Also, one recalls that this was the assumption made in finding the potential of this setup using the Method of Images---the charged plane was assumed to have a charge density $-Q/A$.) Click here to jump to the problem!
 43 Click here to jump to the problem! GR8677 #9 Electromagnetism$\Rightarrow$}Electric Field From symmetry, one has 0-field at the center. This is choice (A). Take the limit of an infinite number of charges placed around the center. In two dimensions, one has a circle of charges. This is similar to a 2-dimensional conductor. There is 0-electric-field inside a conductor (the potential is constant inside). But, for a less-than-infinite number of charges, one would only have 0-field in the center. Gauss' Law inside would have a gaussian surface (just a loop in 2-dimensions) surrounding 0 charge. The brute-force way would be to compute the electric field via Coulomb's Law and vector addition. With the proper coordinates and a $2\pi/5$ angle unit circle, this is a easy calculation. Click here to jump to the problem!
 44 Click here to jump to the problem! GR8677 #10 Electromagnetism$\Rightarrow$}Capacitors The problem gives $C_1=3\mu F$ and $C_2=6 \mu F$ and V=300 V. The capacitors are connected in series, and thus $C_{eq}=C_1C_2/(C_1+C_2)=18/9=2\mu F$. (Recall that series and parallel capacitor formulae are opposite to that of resistors.) Energy is $1/2 C_{eq} V^2 =9E4 \times 1E-6 = 9E-2$, as in choice (A). Click here to jump to the problem!
 45 Click here to jump to the problem! GR8677 #20 Electromagnetism$\Rightarrow$}Bremsstrahlung One might recall the English translation of Bremsstrahlung, which is "braking radiation." The only choice that has to do with acceleration is choice (E). Click here to jump to the problem!
 46 Click here to jump to the problem! GR8677 #38 Electromagnetism$\Rightarrow$}AC Circuit The current is maximized when one has resonance. Resonance occurs when the complex impedance is 0, or when $X_L=X_C \Rightarrow \omega L = 1/(\omega C)$. Plug stuff in to get choice (D). Click here to jump to the problem!
 47 Click here to jump to the problem! GR8677 #39 Electromagnetism$\Rightarrow$}Filters High-pass filters have, for high-frequencies ($\omega \rightarrow \infty$), $V_{in} \approx V_{out}$. The net impedance is given by $Z=R + j(\omega L-1/(\omega C))$. $V_{in}=IZ$, and thus $I=V_{in}/Z$. For cases I and II, $Z=R + j(\omega L)$. For case I: $V_{out}=IR=V_{in}R/(R + j(\omega L))$ Thus, in the regime of high, frequency, one gets $V_{out} \rightarrow 0$ (This is a low-pass filter.) For case II: $V_{out}=IZ_L=V_{in}\omega L/(R + j(\omega L))$, one gets $V_{out} \approx V_{in}$ (To wit: L'Hopital's Rule can be used or this limit.) For cases IV and III, $R + j(-1/(\omega C))$. For Case III, $V_{out}=IR=V_{in}R/(R + j(-1/(\omega C)))$ For high-frequency, $1/(\omega C) \rightarrow 0$, and thus one has $V_{in} \approx V_{out}$. For Case IV, $V_{out}=IZ_C=V_{in}X_C/(R + j(-1/(\omega C)))$. This quantity goes to 0 for high-frequency. (This is a low-pass filter.) Hence, the only choices that work for high-freq filters are choices II and III. Choice (D). Click here to jump to the problem!
 48 Click here to jump to the problem! GR8677 #40 Electromagnetism$\Rightarrow$}LR Circuits Immediately after the switch is closed, the voltage across the inductor is maximal since the change in current is huge. (To wit: $V_L = \dot{I} L$) Thus, only choices (D) and (E) remain. The voltage exponentially decreases rather quickly in a LR circuit, and thus one should choose choice (D) instead of (E). (One can write out the equation $\ddot{Q}L+\dot{Q}R=V$ and solve for V to verify.) Click here to jump to the problem!
 49 Click here to jump to the problem! GR8677 #41 Electromagnetism$\Rightarrow$}Monopoles Theorists like monopoles because if they exist, Maxwell's equation would be symmetrical. Thus, one would get to add terms to the divergence of B and the curl of E to make it symmetric with their other-field counterparts. Choose choice (E). Click here to jump to the problem!
 50 Click here to jump to the problem! GR8677 #42 Electromagnetism$\Rightarrow$}Faraday Law Faraday's law suggests that the current inducted opposes the changing field. (This is also known as Lenz Law). It acts something like an electromagnetic inertia. Since the middle loop is moving towards the observer, the loop A feels an increasing current. Loop A will thus compensate with a current that acts to decrease the field-change; this current is counter-clockwise by the right-hand rule. Since the middle loop is moving away from loop B, loop B will want to increase its magnetic field, and thus its current is in the same direction as the middle loop. Only choice (C) works. (One can immediately cancel out all but choices (A) and (B) from just a conceptual understanding of Lenz Law.) Click here to jump to the problem!
 51 Click here to jump to the problem! GR8677 #58 Electromagnetism$\Rightarrow$}Trajectory The velocity of the proton as it enters the region is given by its acceleration due to the potential difference V. Conservation of energy yields $1/2 mv^2 =qV \Rightarrow v^2 = 2qV/m$. In the first experiment, since the electric and magnetic forces balance, as the particle is un-deflected, one deduces that $qvB=qE \Rightarrow vB = E$ from the Lorentz force. In the second experiment, the potential is doubled to 2V. The velocity is thus greater than it was in the first experiment. Since the magnetic force is proportional to the velocity, one has the magnetic force larger. Thus, the particle is deflected in the direction of the magnetic field, which is in the -x-direction from the right hand rule, as in choice (B). Click here to jump to the problem!
 52 Click here to jump to the problem! GR8677 #59 Electromagnetism$\Rightarrow$}LC Circuit The form for a simple harmonic oscillator is $m\ddot{x} + kx =0$, which one can obtain from Hooke's Law. Comparing this to the LC circuit equation $L\ddot{Q}+Q/C=0$, one sees that $L \Leftrightarrow m$ and $1/C \Leftrightarrow k$ and $Q \Leftrightarrow x$. This is choice (B). Click here to jump to the problem!
 53 Click here to jump to the problem! GR8677 #60 Electromagnetism$\Rightarrow$}Flux Electric flux is given by $\oint \vec{E} \cdot d\vec{A}$. The electric field from the infinite conducting plane is $\sigma/\epsilon_0$. The field points directly perpendicular to the plane of the conductor, and thus the normal area it fluxes-through in the Gaussian surface is just $\pi (R^2-x^2)$. (This geometry is obvious if one draws a right-triangle with the bottom leg of length x, the hypotenuse of length R, and the side-leg bordering the conductor.) Plug everything into the flux equation to get choice (D). Click here to jump to the problem!
 54 Click here to jump to the problem! GR8677 #61 Electromagnetism$\Rightarrow$}Boundary Conditions For a conductor, the electric field boundary condition at the interface are $E_1^\perp - E_2^\perp = \sigma_f$ ($\sigma_f$ is the free charge density) and $E_1^\parallel - E_2^\parallel = 0$. The plane wave traveling in the x-direction is polarized (say) in the +z-direction. Thus, $\vec{E} = \hat{z} E_0 \cos(kx - \omega t)$. There is no component perpendicular to the conductor at the boundary, and thus the first boundary condition implies that the free charge density is 0. The second boundary condition requires that $E_1^\parallel = E_2^\parallel$. The parallel component of E is polarized in the z-direction, and thus the requirement is $E_0 + E_0^{r} = E_{0}^{t}$, i.e., the incident plus the reflected is equal to the transmitted wave. However, for a perfect conductor, the transmitted wave is 0. Thus, one has $E_0 = - E_0^{r}$. This implies that the electric field to the left of the conductor cancels. Recall the Poynting vector, $\vec{S} \propto \vec{E} \times \vec{B}$, which conveniently points in the direction of the electromagnetic wave propagation. Since the electric field (by the convention used above) is polarized in the z-direction, the magnetic field of the incident wave points in the -y direction. However, since the electric field of the reflecting wave points in the -z-direction, its magnetic field also points in the -y direction. The magnetic field magnitude is thus $2E_0/c$. Hence, one has choice (C). Click here to jump to the problem!
 55 Click here to jump to the problem! GR8677 #62 Electromagnetism$\Rightarrow$}Cyclotron Frequency The cyclotron frequency can be easily derived by equating centripetal force with the Lorentz force. It applies to this problem since the magnetic field is perpendicular to the velocity of the particle. $qvB = mv^2/r = mv \omega \Rightarrow \omega = qB/m$, where one recalls the angular frequency $\omega = v/r$. Solving for m in the cyclotron frequency expression above, one has $m=qB/(2\pi f)$, where one recalls that the angular frequency is related to the frequency by $\omega = 2\pi f$. Plug in the numbers to get choice (A). Click here to jump to the problem!
 56 Click here to jump to the problem! GR8677 #85 Electromagnetism$\Rightarrow$}Resistivity This is a baby circuits problem once one calculates the net resistance. Recall that resistance is related to length and area by $R=\rho l/a$. Thus, the resistance of the thin and long wire is $R_{tl}=\rho 2L/A$, while the resistance of the fat and short wire is $R_{fs}=\rho L/(2A)$. The resistors are in series, and so the equivalent resistance is $R_{eq} = \rho L/A(2+1/2)=5\rho L/(2A)$. Plug that into Ohm's Law to find the current $\Delta V = I R_{tot} = 8-1=7$. Thus, $I=14A/(5\rho L)$. To find the potential at the junction, use Ohm's Law again to get $V_{junction}=1+IR_{fs}$. Chunking out the algebraic expressions, one gets 2.4 V, as in choice (A0. Click here to jump to the problem!
 57 Click here to jump to the problem! GR8677 #86 Electromagnetism$\Rightarrow$}Faraday Law The voltage induced is equal to the change in magnetic flux $\epsilon = -\frac{\partial \Phi}{\partial t}$, where $\Phi =\int \vec{B} \cdot d\vec{A}$. Noting the initial condition ($\Phi(t=0)=0$), since the field and area normal are perpendicular), one finds that $B \cdot dA = B \sin(\omega t) \pi r^2$. Thus, $d\Phi/dt = \omega B \cos(\omega t) \pi r^2$. Now, to find the current, one uses Ohm's Law in Faraday's Law to get $IR/N=\dot{\Phi}$, where N is the number of turns. Thus, $I = N/R \dot{\Phi} = NB\omega / R \cos(\omega t) \pi r^2 = 15/2 \times 300/9 \cos(\omega t) \pi (1/100)^2=250E-4\cos(\omega t)$. This is choice (E). Click here to jump to the problem!
 58 Click here to jump to the problem! GR8677 #87 Electromagnetism$\Rightarrow$}Potential Since the potential is constant inside the sphere, one has the field inside to be 0---even though it is not a conductor. Thus, the force contribution comes from the other sphere. By, Gauss' Law, the other sphere acts just like a point-charge, and thus its field contribution is $E=\frac{Q}{4\pi \epsilon_0 (10d-d/2)^2}=\frac{4Q}{4\pi \epsilon_0 361(d)^2}$. The force on the test charge is thus $F=qE = \frac{Q}{\pi \epsilon_0 361(d)^2}$, which is just choice (A). Click here to jump to the problem!
 59 Click here to jump to the problem! GR8677 #95 Electromagnetism$\Rightarrow$}Dielectric No real calculations involved here. Since $E\propto 1/\epsilon$, if one adds in a dielectric then $\epsilon = K\epsilon_0$. Since $E_0\propto 1/\epsilon_0$, this means that $E^{'}=E_0/K$. Click here to jump to the problem!
 60 Click here to jump to the problem! GR8677 #96 Electromagnetism$\Rightarrow$}Symmetry By symmetry, the total power radiated is 0. It expands inwards and outwards at the same frequency. Click here to jump to the problem!
 61 Click here to jump to the problem! GR8677 #9 Electromagnetism$\Rightarrow$}Current Directions The opposite currents cancel each other, and thus the induction (and field) outside is 0. Click here to jump to the problem!
 62 Click here to jump to the problem! GR8677 #10 Electromagnetism$\Rightarrow$}Image Charges The conductor induces image charges -q and -2q since it is grounded at $x=0$. Since these are (mirror) image charges, each charge induced is the same distance from the conducting plane as its positive component. The net force on q is just the magnitude sum of the positive charge 2q and the two induced charges, $\sum F = \frac{q^2}{4\pi\epsilon_0}\left( 1/(a^2) + 2/(2a^2) + 2/(a^2) \right) = \frac{q^2}{4\pi\epsilon_0a^2}\left( 1+1/2+2 \right) = \frac{q^2}{4\pi\epsilon_0a^2}\frac{7}{2}$, as in choice (E). (Why is it the magnitude sum? Well, recall that $\vec{F}=\frac{q_i q_j}{4\pi\epsilon_0 r_{ij}^2}\hat{r_{ij}}$, where $\hat{r_{ij}}$ is the vector pointing from the charge one is interested in to the field charge, i.e., the other charge. If one has two positive charges, then the force points along that unit vector. However, if one has a minus charge and a positive charge, the force points in the other direction. Thus, all the force quantities are additive, and one might as well just take the magnitude sum.) Click here to jump to the problem!
 63 Click here to jump to the problem! GR8677 #11 Electromagnetism$\Rightarrow$}RC Circuit The energy of a capacitor C with voltage V across it is given by $U=\frac{1}{2}CV^2=\frac{Q^2}{2C}$. ($Q=CV$ derives the other variations of the energy.) From Ohm's Law, one arrives at the relation between charge and time, $Q/C+\dot{Q}R=0 \Rightarrow \frac{Q}{RC}=-\frac{dQ}{dt} \Rightarrow -\frac{dt}{RC}=\frac{dQ}{Q}$. Integrating both sides, one finds that $Q(t)=Q_0 e^{-t/(RC)}$. Plugging this into the energy equation above, one has $U \propto Q(t)^2 \propto e^{-2t/(RC)}$. Twice time required for the energy to dissipate by 2 is thus given by $1/2 = e^{-t/(RC)} \Rightarrow t_{1/2}=RCln(2$. Divide it by 2 to get choice (E). Click here to jump to the problem!
 64 Click here to jump to the problem! GR8677 #12 Electromagnetism$\Rightarrow$}Potential A Potential V is related to the electric field E by $\vec{E}=\nabla V$. Since the problem supplies the approximation tool that the planes are quite large, one can assume the field is approximately constant. The remaining parameter that can't be thrown out by this approximation is the angle, and thus the only choice that yields $\frac{d}{d\phi} V = constant$ is choice (B). Click here to jump to the problem!
 65 Click here to jump to the problem! GR8677 #13 Electromagnetism$\Rightarrow$}Maxwell's Equations Magnetic monopoles remain a likeable (even lovable) theoretical construct because of their ability to perfectly symmetrize Maxwell's equations. Since the curl term has an electric current, the other curl term should have a magnetic current. ($\nabla \cdot B \neq 0$ is taken to be obvious in presence of magnetic charge.) The answer is thus (D), and the revised equations are, $\begin{eqnarray} \nabla \times H = J_e+\frac{\partial D}{\partial t}\\ \nabla \times E = -J_m-\frac{\partial B}{\partial t}\\ \nabla \cdot D = \rho_e\\ \nabla \cdot B = \rho_m. \end{eqnarray}$ Click here to jump to the problem!
 66 Click here to jump to the problem! GR8677 #32 Electromagnetism$\Rightarrow$}Circuits Power is related to current and resistance by $P=I^2R$. The resistor that has the most current would be $R_1$ and $R_{eq}$ (the equivalent resistance of all the resistors except for $R_1$), since all the other resistors share a current that is split from the main current running from the battery to $R_1$. Since $R_{eq}, the most power is thus dissipated through $R_1$, as in choice (A). Click here to jump to the problem!
 67 Click here to jump to the problem! GR8677 #33 Electromagnetism$\Rightarrow$}Circuits One can find the voltage across $R_4$ quite easily. The net resistance of all resistors except $R_1$ is $R_{eq}=((1/R_3+1/R_4)^{-1}+R^5)^{-1}+1/R_2=25 \Omega$. Kirchhoff's Loop Law then gives $V=I(R_1+R_{eq})\Rightarrow I=3/75A$. Now that one knows the current, one trivially finds the voltage across $R_2$ to be $IR_2=1$ V. $I^{'}(R_{34}+R_{5})=1$, since the resistors are in parallel. Since $R_{34}=1/R_3+1/R_4=1/60+1/30=20\Omega$, the current $I^{'}=1/(R_{34}+R_5)=1/50$. The voltage across either $R_3$ or $R_4$ is just $1-I^{'}R_5=1-30/50=0.4$, as in choice (A). Click here to jump to the problem!
 68 Click here to jump to the problem! GR8677 #34 Electromagnetism$\Rightarrow$}TEM Waves The full formalism of a conducting cavity can be solved via TEM (transverse electromagnetic) wave guides. However, to solve this problem, one needs only the two boundary conditions from the reflection at a conducting surface, $\Delta E_{\parallel} = 0$ and $\Delta B_{\perp}=0$. The electric field parallel to the cavity is the transverse field, and thus one has choice (D), exactly the conditions above. Click here to jump to the problem!
 69 Click here to jump to the problem! GR8677 #36 Electromagnetism$\Rightarrow$}Boundary Conditions The conductor perfectly reflects the incoming wave, and none is transmitted. The electric field is thus reversed. However, since E and B are perpendicular (related to each other by the Poyting Vector where the direction of propagation is given by the direction of $\vec{E} \times \vec{B}$), the magnitude of B is increased by 2, but its direction stays the same. Search on the GRE Physics Solutions homepage with keyword conductors for more on this. Click here to jump to the problem!
 70 Click here to jump to the problem! GR8677 #54 Electromagnetism$\Rightarrow$}Faraday Law The induced current would act, according to Lenz Law, to oppose the change. In this case, since the field is decreasing (the wire is being pulled away from the field), the induced current would act to increase the field. On the side closest to the long wire, it would thus point in the same direction as the current from the long-wire. This eliminates all but choices (D) and (E). Now, since the rectangular loop wire cannot induce a force on itself, the force is due to the field from the long wire. To the left of the loop, the long wire has a field pointing into the page, and thus the force there is left-wards. One can check again that choice (E) is right by right-hand-ruling the field on the right side of the loop. Since the field due to the long wire is again into the page, the force here is towards the right (since the current runs down the page on the right side of the loop). Click here to jump to the problem!
 71 Click here to jump to the problem! GR8677 #55 Electromagnetism$\Rightarrow$}Magnetic Force The magnetic force of a wire is given by $\vec{F} = I \vec{l} \times \vec{B}$, where I is the current of the wire and l its length. The field that produces the force on the loop is given by the long wire (see the previous problem for why). The field of that wire is given trivially by Ampere's Law to be $\vec{B}=\frac{\mu_0 I}{2\pi r}$, where r is the radial distance away from its center. Only two wires from the loop contribute to the force, since the cross-product yields 0 force for the two horizontal components. Thus, the net force on the loop with current i with vertical components of length b is $|F_{left}+F_{right}|=|ib\frac{\mu_0 I}{2\pi}\left(\frac{1}{r} - \frac{1}{r+a}\right)|$. Combine the fraction to get choice (D). Click here to jump to the problem!
 72 Click here to jump to the problem! GR8677 #57 Electromagnetism$\Rightarrow$}Faraday Law Faraday's Law has the induced voltage is given by the change in magnetic flux, as $V = -\frac{dB \cdot A}{dt}$. (The minus sign shows that the induced voltage opposes the change.) Since the induced voltage has to be periodic (as the half-circle rotates around A), choices (D) and (E) are immediately eliminated. The voltage changes from positive to negative in regions where the change in flux is slowing down, goes to 0, then speeds up again. Thus, choice (C) is out. The change in flux is constantly increasing as the loop spins into the field, and it is constantly decreasing as it spins out of the field. This is choice (A). Click here to jump to the problem!
 73 Click here to jump to the problem! GR8677 #60 Electromagnetism$\Rightarrow$}Cyclotron Frequency The cyclotron frequency is given by $F=qvB = mv^2/r \Rightarrow qB=mv/r = m\omega$, where one merely equates the Lorentz Force with the centripetal force using $v=r\omega$ to relate angular velocity with velocity. So, $\omega = qB/m$. Plug in the quantities to get choice (D). Click here to jump to the problem!
 74 Click here to jump to the problem! GR8677 #64 Electromagnetism$\Rightarrow$}Gauss Law Gauss Law gives $\nabla \cdot \vec{E} = \rho/\epsilon_0$. Since the divergence of E in Cartesian coordinates is non-zero, there is a charge density in the region. QED Click here to jump to the problem!
 75 Click here to jump to the problem! GR8677 #65 Electromagnetism$\Rightarrow$}Small Oscillations The force on the charge in the center due to the charges on both sides is $F=\frac{2Qq}{4\pi \epsilon_0 R^2}$. Small oscillations have a form $\ddot{x} = -\omega^2_0 x$, which comes from $m\ddot{x} = -kx$. Thus, the Coulomb Force above gives $m\ddot{y}=-\frac{2Qqy}{4\pi \epsilon_0 R^3}$. Note the compensating R on the denominator to account for the y. Thus, the angular frequency is given by (E). Click here to jump to the problem!
 76 Click here to jump to the problem! GR8677 #81 Electromagnetism$\Rightarrow$}Resonant Frequency The maximum steady-state amplitude (after transients die out) occurs at the resonant frequency, which is given by setting the impedance of the capacitor and inductor equal $X_C = X_L \Rightarrow \frac{1}{\omega C}=\omega L \Rightarrow \omega^2 =1/(LC)$, as in choice (C). Click here to jump to the problem!
 77 Click here to jump to the problem! GR8677 #83 Electromagnetism$\Rightarrow$}Forces Sum of the forces for one of the mass in the x (horizontal) and y (vertical) directions gives, $\begin{eqnarray} \sum F_x = 0 = T\sin\theta - kq^2/d^2 \\ \sum F_y = 0 = T\cos\theta - mg \end{eqnarray}$ For small angles, $\cos\theta \approx 1 \Rightarrow T \approx mg$. From the geometry, one can deduce that $\sin\theta = (d/2)/L$. Thus, the x equation yields $T(d/2)/L=kq^2/d^2 \Rightarrow d^3 = 2kq^2L/(mg)$ (since $T\approx mg$ from the y equation for small angles). This is choice (A). Click here to jump to the problem!
 78 Click here to jump to the problem! GR8677 #84 Electromagnetism$\Rightarrow$}Accelerating Charges Elimination time: (A) By the Larmor formula, one has $E\propto q^2 a^2$, where q is the charge and a is the acceleration. Since the charge is constant, this choice is true. (B) This is also true by Larmor's formula. (C) True. The energy radiated through a perpendicular unit area is given by the Poynting vector. $\vec{S}\propto \vec{E}\times\vec{B}\propto \vec{E}\times(\hat{r}\times\vec{E}) \propto E^2\hat{r}-(\hat{r}\cdot\vec{E})\vec{E}$ and $E^2\propto 1/r^2$ far away. Also, less rigorously, one can arrive at the same result from recalling the surface area of a sphere, $4\pi r^2$, and thus any term in $\vec{S}$ proportional to $1/r^2$ will yield a finite, thus acceptable answer. (Thanks to the user astro_allison for this pointer. See p460ff of Griffiths, Introduction to Electrodynamics, 3rd Edition for more details.) (D) False. It's a minimum in the plane. (E) True, since far from the electron the field behaves as plane waves, with $E_z$ and $B_z$ both 0. Click here to jump to the problem!
 79 Click here to jump to the problem! GR8677 #88 Electromagnetism$\Rightarrow$}Capacitors From the problem and the basic relation for capacitors $Q=CV$, one immediately deduces that the initial charge is $Q_0=C_0V_0$ and the final charge is $Q_f = \kappa C_0 V_0$, and thus choice (C) is out. The potential is constant $V_f=V_0$, and thus choices (A) and (B) are out. The electric field for a parallel plate capacitor is given by $E=\sigma/\epsilon$. Since $E_0=\sigma_0/\epsilon_0$ and $E_f = \sigma_f/(\kappa \epsilon_0)$, the final field is the same as the initial field. (To wit: $\sigma=Q/A\Rightarrow \sigma\propto Q$.) Thus, (D) is false. (Thanks to the user whose alias is "poop" for pointing this out.) From the definition of $D=\epsilon_0 E_0+P=\epsilon E_0$, one has $D_0, since $\epsilon_0<\kappa\epsilon_0$. Choice (E) is right. Click here to jump to the problem!
 80 Click here to jump to the problem! GR8677 #92 Electromagnetism$\Rightarrow$}Frequency A three-pole magnet should produce three voltage peaks, and thus the frequency is 30 Hz. (Solution due to David Latchman.) Click here to jump to the problem!

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