It must therefore also be zero just inside the boundaries. But there are no surface charges in this setup. Unbalanced current The capacitance of the isolated box must be roughly that of a sphere intermediate between the inscribed and circumscribed spheres. Automobile battery If the voltage drop across the 0.
The voltage drop across the internal resistor is then Ri But we know that this voltage drop is Therefore, 2. In the second box, in each case we have one resistor in parallel with the series combination of the other two resistors. And in the second box the ratio is 34 to 85, which is the same.
The two boxes are therefore indistinguishable by external measurements using direct currents. Now consider the tapered cone. If b a then this fraction is larger or smaller than 1, which makes sense. See Problem 4. As a check on this result, consider three objects, all with the same length: a cylinder with radius a, a tapered cone with radii a and b, and another cylinder with radius b. For concreteness, assume a Or you can just look at Eqs. This result is independent of n.
This result is independent of f. If there are n resistors connected to each node n was 4 in Problem 4. All of the setups have the necessary symmetry for the current to divide equally.
All the other resistors outside the two nodes are irrelevant. Resistances in a cube a In Fig. So the circuit is equivalent to the second setup shown in Fig. Alternatively, we can work in terms of currents.
These vertices are all at the same potential halfway between VA and VB , so we can collapse them to a point. In the second setup shown, there are only 10 lines because 2 of the original 12 lines were collapsed. As expected, this is smaller than the answer to part b , which in turn is smaller than the answer to part a.
This is a special case of the general result in Problem 4. This current splits into the currents I1 and I2. These results make intuitive sense. For example, after one step of the ladder, we would have the setup in Fig. Let the desired resistance be r. Following the strategy from Exercise 4. The circuit to the right of C and D in Fig. You should convince yourself why this setup is actually the same as the setup in Problem 4. So r2 should equal the parallel combination of 1 and r1.
This works out due to the various properties of the golden ratio. Both bulbs have the same voltage drop V , so if Bulb 1 is twice as bright as Bulb 2, it must have half the R. The larger resistor is dimmer. Both bulbs now have the same current I, so if Bulb 2 has twice the resistance, as we found in part a , then it is twice as bright — the opposite of the case in part a.
The larger resistor is brighter. We can also compare the total power dissipated in each case. This makes sense; we want the largest possible current passing through the given external resistor. D-cell a The total charge produced by 0. Since the mass of the battery is 0. In the example in Section 4. This is about half as much as the D cell. This corresponds to about But by looking at the bottom loop in the circuit, this means that R2 must be very large, roughly 1. In particular, there is no way it can cut the current in half.
Our initial assumption that R1 is not much smaller than Ra must therefore have been incorrect. If you want to solve the problem exactly, you can show that R2 should be And R1 should be 0. Using symmetry The symmetry is evident in Fig. The second missing term in the numerator must be R1 R4 the mirror-image of the R2 R3 term.
The missing term in the denominator must be R2 R3 the mirror-image of the R1 R4 term. The parallel combination of R1 and R3 is in series with the parallel combination of R2 and R4. The series combination of R1 and R2 is in parallel with the series combination of R3 and R4. This agrees with the formula when all terms containing R1 or R3 are dropped. We then have the circuit shown in Fig.
And likewise for R2 and R4. So the total resistance is R. You should check that the formula agrees with what you calculate directly. You should verify that you obtain the same result if you simply connect A and B in Fig. In the Thevenin equivalent circuit, Eeq is simply the open-circuit voltage, 0. And Req is given by Eeq 0. The resistance Req can be found by ignoring that is, shorting the emf.
From Exercise 4. Alternatively, we can solve the problem from scratch. Discharging a capacitor From Eq. Suppose we have a 1 microfarad capacitor charged to volts. Since the charge of an electron is 1. From the solution to Problem 4. Remark: It is also possible to use the general formulas for Q t and I t from Problem 4. The NS dimension is unchanged, as is the vertical separation. The number of electrons on the negative plate is the same. The number of electrons is the same. Electron beam a 0. Note that Ex is very close to zero.
So there must be an angle in between where the factor equals 1. Very thin! See Fig. This follows from the fact that in Fig. Figure 5. A nonuniformly moving electron a The electron had been traveling in the positive direction along the negative x axis toward the origin, where it rather suddenly stopped. The thickness of the shell containing the connecting lines is determined by the duration of the deceleration period.
This thickness remains constant. This radiation is called Bremsstrahlung. This is demonstrated in Appendix H, but we can also understand it in another more qualitative way. We have used the fact that the thickness of the shell remains constant. We could plug in the numbers, but since we want to write the result in units of mc anyway, it is easier to take the following route.
And the V in Eq. The other two dimensions are unchanged. If you want, you can show that taking relativity into account would decrease the speed by only about 0.
Let P2y be the y component of the momentum of particle q2. So P2 points in the y direction, that is, it is perpendicular to v. But we can also calculate this in the same manner as above, by considering a cylinder with radius b, drawn with the path of q2 as its axis; see Fig.
The E2y here is the magnitude of the y component. Consider q2. For nonrelativistic motion, this says that the kinetic energy of q2 must be much larger than the electrical potential energy of the particles at closest approach. It is reasonable that the condition takes this form, because these are the only two energy scales in the problem.
Decreasing velocity Force equals the rate of change of momentum. Therefore, since there is no force in the x direction in the lab frame, px must be constant. Therefore, vx must actually decrease. We can also demonstrate this result by looking directly at the forces, without mentioning momentum. This can be seen as follows. The x speed in the lab frame therefore decreases via the velocity addition formula. Stationary rod and moving charge a The force is always largest in the rest frame of the particle.
We therefore need only worry about the electric force. In this frame, Eq. Remember, the force is always largest in the rest frame of the particle on which it acts. The task now is to explain the force as the sum of the electric and magnetic forces in the lab frame.
As mentioned in the problem, Eq. This is consistent with the Lorentz transformations in Chapter 6. Note that the electric and magnetic forces in the lab frame have exactly the same magnitudes that they had in the second example in Section 5. Moving perpendicular to a wire In the notation of Fig. So for the left electron in Fig.
We now have a handle on all the necessary quantities, so we can use Eq. The denominators are therefore equal at both limits. Since v is constant because the magnetic force is always perpendicular to the velocity , we see that R is constant, which means that the path is a circle. Second solution: We can also calculate R in a more mathematical way.
Assume that B is uniform. The x and y components of Eq. Proton in space The speed is essentially c, so from Exercise 6. The z component of B produces a force in the plane of the loop, which contributes nothing to the torque around the x axis or the y axis and can therefore be ignored.
As an exercise, you can also show that it produces no torque around the z axis. But loop y dx equals the area a of the loop. This is true because in Fig. The sum of these signed areas is the area of the rectangle from B to A. Adding up all such rectangles gives the area a of the whole loop.
This works even if B is below the x axis; y is now negative, so the areas add, which again yields the entire area of the thin rectangle. The loop will therefore tend to spin around an axis pointing in the x direction. So Ny involves the integral x dx around the loop, which you can quickly show is zero; the upper and lower parts of the loop now exactly cancel instead of only partially canceling.
The net force is therefore zero. Determining c If we follow the wires around in Fig. The downward electric force on the upper plate can be obtained in various ways. From Section 3. The alternating voltage allows there to be except at discrete moments during each cycle both a nonzero charge on the capacitor C1 and a nonzero current in the rings.
If the two current densities are equal and opposite, then there will be zero current in the hole, in agreement with the given setup. Given the ratio of the areas of the two circular cross sections, currents of A into the page and A out of the page will yield the given A into the page.
A more remarkable fact see Exercise 6. The current densities will then cancel in the region of the smaller rod, creating the desired cavity.
Consider a point inside the cavity, at position r1 with respect to the center of the given rod, and at position r2 with respect to the center of the cavity, as shown in Fig. It is perpendicular to a and proportional to a. And even more generally, adding on any vector with zero curl also works. Comparing this with the B in Eq. You might be wondering how this can be, in view of the in the positive z fact that Eq. The answer is that we can add an arbitrary constant to the A in Eq. We can therefore ignore the return semicircular path.
The currents along six of the edges cancel, and we end up with the desired square loop of current. If r is the distance from a given point P to the small piece, then Fig. The dl for the larger piece is twice the dl for the smaller piece. This reasoning works for any ratio of ring sizes, of course. Field at the center of a disk Consider a ring with radius r and thickness dr. It points perpendicular to the disk, with the direction determined by the righthand rule.
The contributions do indeed add and not cancel. For comparison, the peak current in a bolt of lightning is on the order of A. Right-angled wire Consider the contribution from the positive x-axis part of the wire. In the notation of Fig. We therefore end up with the desired result. Force between a wire and a loop Consider a little segment in the right-hand side of the square. The vertical components of the preceding two forces cancel, but the leftward components add. So the net force is leftward, as desired.
You can quickly show that the net force on each of the other two sides of the square is zero. In contrast, the horizontal component of B does not change sign, so the negation of the current causes a negation of the vertical force. The net vertical force is therefore zero. The above reasoning shows where the two factors of R in the denominator come from.
It will rotate counterclockwise when viewed from the side. This is consistent with conservation of angular momentum, because the straight wire will gain angular momentum relative to, say, an origin chosen to be the center of the square as it moves to the right. This angular momentum will have a clockwise sense, consistent with the fact that the total angular momentum of the system remains constant.
If the currents in the rings are equal and point in the same direction, then from Eq. The deviations will then be of order z 4. A continuity argument shows why there must exist a point where the second derivative of Bz z equals zero. Field at the tip of a cone Consider a circular strip with width dx, a slant-distance x from the tip. Note that the result in Eq. A rotating cylinder Eq. What is the current density in our rotating cylinder? We can do the same with the other patch in the same cone.
If the projected patches are distances r1 and r2 from the point P , then their areas are proportional to r12 and r22 , because areas are proportional to length squared, and because the patches cut the line from P at the same angle perpendicular, by construction.
The numerators in the Biot-Savart law for the corresponding patches are therefore proportional to r12 and r These factors exactly cancel the r2 in the denominator of the Biot-Savart law. So the magnitudes of the contributions from the two patches equal. But there will still be equal numbers of patches having currents in each direction which can be traced to the fact that P has the property of being outside the solenoid , so the sum of the contributions will still be zero.
Rectangular torus Loop 1 Loop 2 B1 I B2 Figure I Consider two loops of current that are located symmetrically with respect to a plane through the axis of the torus; a top view is shown in Fig. You can check this by looking at the Biot-Savart contributions from corresponding little pieces of the two loops; the components parallel to the plane cancel.
In general, the B1 and B2 vectors shown also have components perpendicular to the page, but you can show that these components are equal and opposite. The same Biot-Savart reasoning involving corresponding little pieces holds. The entire coil can be decomposed into pairs of loops located symmetrically with respect to a given plane. This expression for B holds for a torus of any uniform cross section. So the deviation is about 2. We expect the departure from uniformity in the radial direction to be of the same magnitude, roughly, as the variation in the axial direction.
A less elegant way of solving this exercise is to use Eq. In other words, it was horizontal, as we wanted to show. Of course, this holds only if r0 6. Equal magnitudes This setup could be created by taking the setup in Fig.
So they are perpendicular, as desired. But the magnitudes B1 and B2 are equal, so the force is zero. So the force on the moving charges is zero. In general, they will simply make equal angles with the direction of the current.
There will then be a nonzero force on the sheet. Force on electrons moving together a Since there is no transverse length contraction, the distance between the electrons in the frame in which they are at rest is still r. To transform this force to the lab frame, we can use the fact that the transverse force on a particle is always greatest in the rest frame of the particle. This follows from Eq. Alternatively, it follows from the transformations in Eq.
The transformations in Eq. Bottom-stick frame: In this frame, the situation is shown in Fig. The charge q and the top stick are now moving. See Problem 6. The right end of the charged region is determined by the right end of the disk, which is at radius That is what Rowland did. The Hall voltage across the ribbon of width 0. Oscillating E and B Consider a circle of radius r centered on the axis.
Imagine connecting the ends of the wire with another wire to form a complete loop, which is in fact what you would be doing if you measured the voltage between the ends by connecting them to a voltmeter. We can also solve this exercise by looking at the qvB magnetic force on the charges in the wire.
For any particular F that you pick, Eq. We can now answer the various questions. So a force of 2 N will pull the frame out in half the time, or 0. So a brass frame will be pulled out in 1 sec by a force of 0. So a 1 cm aluminum frame will be pulled out in 1 sec by a force of 4 N. Sliding bar a Let v be the instantaneous velocity of the bar.
We see that the velocity decreases exponentially, so technically the rod never stops moving in an ideal world. This exponential decay of v is a familiar result for forces that are proportional to the negative of v. A loop with two surfaces Surface a has two sides, whereas surface b has only one side; it is a Mobius strip.
So it can be considered a two-turn coil. The three-turn coil in shown in Fig. If you trace along the wire as it spirals upward, it takes the same shape as the windings of a solenoid. The pancakes are truly all part of a single surface. We therefore obtain the result that a coil of N turns has an emf that is N times that of a single loop. The current is therefore clockwise when viewed from above.
We can alternatively write the condition in Eq. To see how this works out in a simple case, let the coil have the same rectangular shape as the loop, but with N turns. And let the coil have current I0. To see why, imagine slicing the loop into strips oriented perpendicular to the axis of rotation. This is true for the following reason. Consider a nonplanar surface bounded by the loop, and divide it into many tiny planar patches.
This is easily seen in an extreme case where B is zero except in a given region; see Fig. The emf will be zero except when the loop is sweeping through that region, so the emf will look something like the curve in Fig. The nonzero parts of the curve belong to a sine curve. The numbers shown correspond to the orientations in Fig.
The former fact will lessen the peak in the standard sinusoidal curve, and the latter fact will increase E near its zeros. Cases a and c are equivalent for path 2, but not for path 1.
The total voltage drop around a closed path is zero in cases b and c consistent with the fact that these are electrostatic setups , but nonzero in case a. A little element of charge dq feels a tangential force E dq, and hence a torque Ea dq.
So you can quickly show that if q is positive, the direction of L is the same as the direction of B0. We see that L depends only on the net change in B, and not on the rate at which this change comes about.
Intuitively, if B changes more slowly, then E and hence the torque is smaller, so the angular momentum increases at a lesser rate. But the process takes longer, so the increase goes on for a longer time.
This is due to the fact that E is proportional to a, which means that the linear acceleration in the tangential direction, and hence the tangential speed v at a given time, is proportional to a. From Table 4. Suppose the coil is cylindrical, with length h and radius a. We have another clue: the two coils are wound closely together separated only by twine.
The winding must look something like what is shown in Fig. So the coil is about a foot long and a foot in diameter — very reasonable. The reconstructed circuit is shown in Fig. Then with the switch closed, the steady current through coil 1 is 50 amps. The current pulse through the galvanometer therefore looks something like the curve in Fig. But the existence of coil 2 allows the current in coil 1 to stop. M for two rings From Eq. In this limit we can also neglect the variation of B over the interior of the other ring.
Connecting two circuits a In Fig. The larger the outer ring is, the smaller this central region is. Net flux through outer ring comes from this central region ring 1 ring 2 Figure 7. Using the mutual inductance for two rings With current I1 in the outer ring, Eq. We will have more to say about this form of B in Chapter Of course, the inductance is not exactly zero. But real wires have thickness, so the self-inductance of the hair pin will indeed be small. Rather, the area spanned by the wire in Fig.
This area has nothing to do with the inside of the cylinder. This means that we have overestimated the inductance; the true value is smaller than the above approximate result. Opening a switch After the switch has been closed a while, the currents are steady and the inductor is irrelevant. So 10 V is the initial voltage across each of the branches of the circuit. They are both directed downward. Initially both A and B are at 10 V with respect to ground. Right after the switch is opened, we have the circuit shown in Fig.
Therefore, the current through the circuit is 0. After 0. The plots are shown in Fig. Note the discontinuity in VA. Figure 7. RL circuit From Eq. The current reaches a value of 0. The task of Problem 7. Decay time for current in the earth From Eq.
A dynamo The device on the bottom is the dynamo. Positive charges in the disk will be pushed outward, negative charges pushed inward. This conclusion is independent of the sign of the mobile charges. A dynamo of this kind runs equally well with current in either direction. The current can also be zero. It retains the direction of its initial excitation.
The magnitude of the current in this purely ohmic dynamo is determined by the input mechanical power. Chapter 8 Alternating-current circuits Solutions manual for Electricity and Magnetism, 3rd edition, E. So the voltage across the inductor must also be zero. But from Eq. During this period we therefore essentially have a series RLC circuit in the right loop.
There are two things we can estimate by looking at the given plot: the frequency of oscillation and the rate of decay of the signal. These two things will allow us to calculate C and R, respectively. Therefore, 2L 2L 2 0. No current passes through the capacitor in the eventual direct-current steady state. And there is no voltage drop across the inductor for constant current. The 40 in the denominator is of course inconsequential. Resonant cavity As indicated in Fig.
From Exercise 6. The charge oscillates back and forth. Solving an RLC circuit Let I1 and I2 be the loop currents in the left and right loops, respectively, with clockwise taken to be positive.
Let V be the voltage across the capacitor, taken to be positive when the upper plate of the capacitor is positive. So we can quickly obtain the results in Eq. Using Eqs. Overdamped oscillator From Problem 8. Note that both roots are positive, as they must be, because a voltage that grows with time would violate conservation of energy. In the setup in Fig. In the underdamped case, the voltage is given in Eq. For large t, the critically-damped energy in Eq.
Note that the capacitor and inductor have the same energy in the limit of large t. Intuitively, it is believable that the energy decay is quickest for critical damping, because the decay is very slow at both extremes of very light damping and very heavy damping. So the maximum decay rate must occur somewhere in between, and critical damping is as good a guess as any.
And large damping means large R, so there is essentially no current, making the I 2 R term negligible; the energy slowly leaks out of the capacitor.
There is no If I t which is the integral of I t , equals Ie need for a constant of integration because we know that Q t oscillates around zero. So we simply have a resistor in series with the voltage source. In this case, the charge which has a maximum value of CE0 on the capacitor sloshes back and forth very slowly, so the current is very small.
So we simply have a capacitor in series with the voltage source. The current is ahead of the voltage, because the current reaches its maximum while charge is building up on the capacitor, and then a quarter cycle later the charge reaches its maximum.
We are taking Q to be the charge on the top plate of the capacitor, as we did in Section 8. We want to have the same current, 0. So we quickly see that the third curve must be the applied E.
So they must correspond to the 2nd and 4th curves. So the 4th is L, the 1st is R, and the 2nd is C. The order of the curves is therefore R, C, E, L.
Note that E is slightly ahead of R or equivalently I. So I is behind E. So from Eq. That is, the impedance of the inductor is larger than the impedance of the capacitor. Z The frequency is low enough so that the inductor lets current through easily compared with the resistor and capacitor.
The frequency is high enough so that the capacitor lets current through easily compared with the resistor and inductor. Small impedance We have a capacitor in series with the parallel combination of a resistor and an inductor. The system is on resonance. So we have the counterintuitive result that decreasing the resistance in the circuit increases the impedance.
This case has the largest possible Z. Real impedance We have an inductor in series with the parallel combination of a resistor and a capacitor. Equal impedance? In the right equation, the negative sign implies that the only way the relation can be true is if both sides are zero. Note: if you solve this exercise by multiplying Eq.
A spurious solution is introduced, and a valid solution is missed. This makes things more complicated. Finding L The setup is shown in Fig. If we then measure the voltage across the inductor and obtain Consider the general circuits shown in Fig. If the circuits are equivalent at every frequency, they must behave the same way for any pulse or transient.
Consider a battery applied across the terminals which charges the capacitor in each circuit. This is true for the given circuits. So the pattern will be an ellipse, as shown in Fig.
If the plates are connected in the natural way as shown, then the ellipse is traced out counterclockwise. And the current is zero at this moment, so the voltage across the resistor is zero. We see that the curve on the screen passes point B a quarter cycle after point A. So the curve is traced out counterclockwise. On the other hand, if the connections are made in the reverse manner for either of the elements, then the curve would be traced out clockwise.
If both connections are reversed, then the trace reverts back to counterclockwise. Without being told which way the connections are made, there is no way to know the direction of the trace. The power is proportional to V 2. The physical reason why V1 decreases with decreasing frequency is the following. For very high frequency, the impedance of the inductor is very large.
The impedance of the resistor is negligible in comparison, so essentially all of the V0 voltage drop occurs across the inductor, which is what V1 registers. On the other hand, for very low frequency, the impedance of the inductor is very small; it is essentially a short circuit. Therefore, essentially all of the V0 voltage drop occurs across the resistor. Very little occurs across the inductor which, again, is what V1 registers. As in Problem 8. But this voltage is simply E0 because we have a parallel circuit.
The resistors are the only places where power is dissipated. The complex voltage across the right resistor is simply E0. But since we this direction and to E. The other direction would simply change the sign of E0 ; the sign is arbitrary, since the trig function switches signs anyway.
Kicked by a wave Equation 9. So as mentioned in the statement of the exercise, the magnetic force is indeed negligible. The wave tends to knock the particle along. This equals c, as we know it should. You can show with a Taylor series that these level curves are circles near the origin. They are shown roughly in Fig. Since E has only a z component, it points perpendicular to the page.
Small values of C yield near-squares close to the boundary of the box, and values close to 1 yield the small near-circles close to the origin we found above. Microwave background radiation As shown in Section 9. However, the power is undoubtedly emitted in at least a somewhat directed manner, so the distance from an actual radio transmitter would be larger than this. As in Section 9. So the direction is correct; the energy in the wire increases, consistent with the fact that it heats up.
Since J points upward in Fig. So it points into the page on the right. See the remark at the end of the solution to Problem 9. As mentioned at the end of the example in Section 9. As in Problem 9. And Eq. At the Problem 9. Alternatively, you can use the integral table in Appendix K. The sum of Eqs. It is possible for both densities to decrease while the total charge remains at the given value Q, because the charge in the right region increases while the charge in the left region decreases , but it does so at a slower rate than the area increases; so the density decreases.
We would have a paradox if the areas stayed the same. An analogy: 10 people each have the same amount of money. The total amount of money held by all 30 people is still the same, but the average amounts in the two groups now with 11 and 19 people have both decreased. Leyden jar Assume that the jar is cylindrical, with the height being twice the diameter d the result will depend somewhat on the proportions assumed.
So the capacitance is s 2 C2 2 4 8. The diameter is then However, the capacitor can deliver all the stored energy in less than a microsecond! Problem 3. Capacitor roll Let w and s stand for the width and thickness of the materials. This would ruin the capacitor, because we need the two strips to have opposite charge. So we need to add on a second layer of polyethylene, as shown in Fig.
So it appears that the total length of each kind of tape should be 2 1. However, from Problem 3. To calculate the diameter of the rolled-up capacitor, note that the area in the plane of the page of the capacitor in Fig. The given distance of 3 m is reasonably large compared with this, so our dipole approximation is a fairly good one. However, if s or C were increased enough, then the length scale of the capacitor would be on the order of 3 m.
So we are interested in two points in the yz plane like the ones shown in 2 Fig. Field lines Let the dipole point in the z direction. Alternatively, we can work with polar coordinates, as we did in Section 2. This holds for any vertical plane containing the z axis, so it holds for Ey too. Hence the averages of both Ex and Ey over the whole surface of a sphere are zero. Using the form of Ez given in Eq.
All four charges have either x1 or x3 equal to zero, so the x1 x3 terms in the matrix vanish. Also, all of the charges have the same value of r2 , so the sum of r2 over all the charges is zero, because the sum of the charges itself is zero. We are therefore left with only the 3x21 and 3x23 terms. This is demonstrated in Fig. Force on a dipole Let the middle dipole be located at the origin. The left end therefore feels a larger force upward than the right end feels downward.
So the total force on the middle dipole is upward and rightward. You can be quantitative about this if you want. The task of Problem Mutually induced dipoles Consider the setup shown in Fig. Field from hydrogen chloride From Eq. So we have the distribution shown in Fig. Let x be the distance from the chlorine nucleus to the electron center. This would be the case if the electron from the hydrogen atom remained spherically symmetric around the hydrogen nucleus the proton.
The values of kT at these temperatures are 4. Equivalently, since the proton mass is 1. Finally, to obtain the number of molecules per m3 , N , we must multiply by The dipole moments, p, from Fig.
This component points inward in the upper hemisphere and outward in the lower hemisphere , because E points downward. This component points outward in the upper hemisphere and inward in the lower hemisphere. The direction is downward. This is the numerator of the result in Eq. Remark: A similar result holds if we kick things down another dimension and consider a slab.
The same reasoning holds again, with two slightly displaced slabs. Conducting-sphere limit Equation The sphere therefore becomes an equipotential, which is correct for a conducting sphere.
This is very close to the Bohr radius, 5. Figure Both E and P are zero outside the slab, so the external D is likewise zero. Our task is therefore to show that D is zero inside the slab.
To see physically why the energy is larger, consider the case of induced dipole moments, discussed in Section The E and B energy densities are therefore equal, just as they are in vacuum. Earth dipole a Equation Sphere dipole Let the axis of rotation be vertical. Also, from Exercise It makes sense that this factor is larger than 1, because if the shell is projected onto the equatorial disk, the density near the rim is larger than the density at the center.
On this assumption, Eq. The physicists are feet away, which is about 30 meters. This is times the 0. If it were perfectly steady it would not be noticed. Field somewhat close to a solenoid Consider a single loop of the solenoid at its middle. Due to the reciprocity theorem, this is also the M going the other way. I1 and B1 are technically functions of time. But since they have the same time dependence, we can take them to be the amplitudes a vector amplitude in the B1 case.
But the latter produces no force on the current in the wire, so we care only about the z component. Since the current in the wire in Fig. Equation The rightward force on a little piece dx of the wire equals I1 Bz dx. See the reasoning in the paragraph following Eq. This force is consistent with the magnitude of the leftward force on the square loop we found in Eq.
Dipoles on a chessboard a Equation The line integral of the gradient of a quantity is simply the change in that quantity. That is, we expect all of the dipoles to be bound. The distance r between the various squares is found from the Pythagorean theorem.
These entries are repeated in other parts of the board; there are only ten independent entries. However, these are only negligibly more bound than the other interior dipoles. So the magnetic force on each of our magnetic dipoles is conservative, just as the electric force on an electric dipole is. So we can use the same reasoning we used back in Chapters 1 and 2 to say that the total work required to remove all of the dipoles far from 1. You can check that the negative sign out front makes the overall sign correct.
Alternatively, if we look at, say, m1 , then Eq. So there is a force in the y direction. Alternatively again, there is no need to even mention forces. So no work is done. The work required to rotate a dipole is the integral of the torque with respect to angle. The sign is correct because the work is positive; we are making m1 be more antiparallel to B2. However, as you can check, it is also zero in the analogous cases of a tetrahedron, a cube with the dipoles aligned parallel to an edge, and a cube with the dipoles aligned parallel to a long diagonal this case involves some tricky angles.
So it is reasonable to conjecture that the potential energy is zero in the case of any platonic solid, for any common orientation of the dipoles. Note that our setup with magnetic dipoles is equivalent to one with electric dipoles, because the forces and torques take the same form in electric and magnetic dipoles since there are no external currents involved; see the discussion following Eq. You can show that the potential energy of this system is zero. Electric vs. Please read our short guide how to send a book to Kindle.
Save for later. This site is like a library, you could find Edward Purcell Electricity and Magnetism. For 50 years, Edward M. Purcells classic textbook has introduced students to the world of electricity and magnetism.
Free edition of Purcell, Electricity and Magnetism. At a link from the. In he shared the Nobel Prize for Physics for the dis-covery of nuclear.
Electricity and magnetism Edward M. Morin, Harvard University. Purcell Solutions Manual. Ebook PDF. PDF download. This site is like a library, you could find million book here by using search box in the header. If searching for a ebook Solution manual berkeley.
Electricity and magnetism purcell 3rd edition solutions pdf. Is a loose url forwarding carrier url redirection allowing everybody to take any present url Download Electricity And Magnetism Purcell Solutions Manual book pdf free download link or read online here in PDF.
This site is like a library, you could find Manual Download Smartphysics electricity and magnetism manual solutions. The lab Edward purcell solution manual — electricity and magnetism Get electricity and magnetism purcell solutions manual PDF file for free from our online libr. We advise you to search our broad variety of pdf of which distribute from many Purcell Solutions Manual. Device electronics for integrated circuits 3rd edition solution manual View solution-manual-device-electronics-for-integrated-circuits-3rd-edition-richard-s from ECON at Harvard University.
Full file at College physics 8th edition serway solutions manual pdf Get instant access to our step-by-step College Physics solutions manual. Our solution Cengage stewart calculus student solutions manual Student Solutions Manual — Provides completely worked-out solutions to all odd-numbered exercises in Skip to the content Home November 13 Electricity and magnetism purcell solutions manual pdf.
November 13, December 13, Previous post Dynamics 12th edition solution manual. Next post Electrochemical methods solutions manual pdf. January 9, January 6, January 1,
0コメント