Physics Practice Problems

The PH212 practice problems have been moved to yhwhan.googlepages.com/physics2 (they include rotational kinematics, special relativity, oscillations, waves, sound, and light).

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A. Somebody wanted me to put up my own solution for problem 42 in the Chapter 28 homework, and I can see why, since Dennis’s solution says "We use the result from Problem 36..." and nobody knows where the solution to 36 is.

So the problem is (equivalent to) this: Suppose you are looking down at a square loop of wire lying flat on the table, with a current I going counterclockwise on it. What is the magnetic field at a point a distance x directly above (out of the table) the center of the square loop? The length of each side is L.

The solution isn’t too bad, but it’s certainly not simple. The main thing is the Biot-Savart Law: dB=μ0I dl·sin θ/(4πr²). Note that here, the scalar form is preferable to the vector form because r^ is neither perpendicular nor parallel to dl.

B=0I L²/4π(x²+L²/4)(4x²+2L²)

Solution, Part 1 (just to get you started):

So, we start by splitting it into four equivalent pieces: The field due to each side of the square is the same, and each has a length L. Also, lets create a coordinate system so that the center of the square is at the origin. Looking down at the square loop and at the coordinate system, the i axis will be pointing to the right, the j axis will be pointing to up, and the k axis will be pointing outwards, straight at us (and the point we want to find the magnetic field at is on this axis). Also, to avoid saying "the point where we want to find the strength of the magnetic field at" every time we want to talk about the point where we want to find the strength of the magnetic field at, I am going to very creatively name that point P.

Well, we will have to choose a side to integrate over and multiply by four, so I am arbitrarily going to choose the bottom side. This way, dl=dli, since the current is going towards the right in this segment of the wire. And if we are going to integrate over this side, we might as well create a sort of mini coordinate system, so let’s have l, the variable we’re going to integrate over, be 0 at the center of the segment, negative to the left of this center, and positive to the right. Then (you’ll want to sketch a diagram of this one) r will be pointed out (at us) x, up L/2, and to the right -l (or to the left l), and making r=xk+½Lj-li. Note that ri is negative l, because when l is positive, the dl segment of wire is to the right of the center, and P is at the center, so r^ would point to the left—i.e., in the -i direction—at P.

As for θ, the angle between dl and r? Well, imagine we draw a triangle, starting from the center of the bottom side (l=0), to the left along the wire a distance l til we get to dl (l=-l), making a corner there at dl and continuing along r til we get to P, then make another corner and go straight back to the center of the bottom side, where we started. The third leg and the first leg are perpendicular, yes? And the first leg is a length -l, yes? So, if a=length of third leg, tan θ=a/-l=-a/l. By the way, notice that as a is the length from P to l=0 (the center of the wire segment), we already know (via Pythagorean’s theorem) that a=(x²+L²/4).

Solution, Part 2 (everything but spelling out the integral):

Well. We could just use plug in arctan(-a/l) for θ, but how will we ever integrate that? (Those of you taking Math 565 or something, be my guest). Better, instead, to observe that this implies that l=-a/tan θ=-a·cot θ, which further implies that (after you differentiate) dl=-a·-csc²θ·dθ=a·dθ/sin²θ. Wait—isn’t sin θ=a/r ? So dl=a·dθ/(a²/r²)=r²dθ/a !

Ok, so why does this deserve the highly prestigious Exclamation Mark Award? Don’t forget the Biot-Savart Law: dB=μ0I dl·sin θ/(4πr²). If we plug in the above for dl, then we get dB=μ0I r²dθ·sin θ/4πr²a=μ0I dθ·sin θ/a=μ0I·dθ·sin θ/4πa, where we can cancel the r’s!

And now that we’re changing the variable, we might want to start paying attention to what this will do to our limits of integrations. More specifically, it was originally from -L/2 to L/2, and now what is it from? Draw the diagram. We want to integrate θ, the angle between dl, which points to the right, and r, which points to P, from an acute angle when dl is at far left end of this segment of wire (l=-L/2), through a right angle as dl passes through the magical l=0 point, to an obtuse angle when dl reaches the far right end (l=L/2). For the lower limit of integration, clearly θ=arctan(-a/(-L/2))=arctan(2a/L). But for the upper limit of integration, when dl is on the right, and l=L/2, we want θ>π/2, but arctan(-a/L/2)=arctan(-2a/L)<π/2, in fact, arctan(-2a/L)<0. Remember your trigonometry? There is actually more than one possible θ for any particular tan θ=x, and to get every θ then actually θ=arctan(x)+k·π, where k is any integer at all. The θ we want now, rather than k=0 as before, is when k=1, so θ=arctan(-2a/L)+π.

Solution, Part 3 (if you do need me to spell out the integral):

So. ∫dB=μ0I/(4πa)arctan(2a/L)arctan(-2a/L)+πsin θ·dθ=μ0I/4πa·(-cos θ|arctan(2a/L)arctan(-2a/L)+π)=μ0I(cos(arctan(2a/L)) - cos(arctan(-2a/L)+π))/4πa. How do we find cos(arctan(2a/L))(and cos(arctan(-2a/L)+π)) ? Well, for the first case, the easiest to understand method would be to say, heck, this is equivalent to saying we want to find cos θ where θ=arctan(2a/L) => tan θ=2a/L. Well, the latter implies that this would be valid for a triangle where the side opposite θ is of length 2a and the side adjacent to it is of length L, in which case the hypotenuse would be of length (4a²+L²), which in turn implies that, cos θ=adj/hyp=L/√(4a²+L²). Alternatively, a perhaps more mathematically rigorous way would be to use the two trigonometric identities tan θ=sin θ/cos θ and sin²θ+cos²θ=1: we know tan θ (it’s 2a/L), so two equations and two unknowns (sin θ and cos θ), should be no problem to solve. The former isn’t so easily applied to arctan(-2a/L)+π, but the latter is (if you remember to also use cos(A-B)=cosAcosB+sinAsinB, which results in -L/√(4a²+L²) ).

Solution, Part 4 (what to do after you’ve solved the integral):

Anyhow, so far we’ve got that B due to the bottom side of the square=μ0I·(L/(4a²+L²)--L/(4a²+L²))/4πa=μ0I(2L/(4a²+L²))/4πa=μ0I L/2πa(4a²+L²)=μ0I L/(x²+L²/4)√(4√(x²+L²/4)²+L²)=μ0I L/(x²+L²/4)√((4x²+4L²/4)+L²)=μ0I L/(x²+L²/4)(4x²+2L²). But wait, this doesn’t take into account the fact that part of this is canceled out, right? Because the B due to each side is not directed straight out of the table, part is directed off to the the side a bit, and these side components cancel each other out when you add up each one corresponding to each side of the square. So which part of B found above do we actually multiply by 4 to get Btotal? Let’s draw a new diagram, focusing on the one segment of wire and ignoring the rest. Let’s pretend we’re looking at the experiment from the side down the length of the wire, so the wire is just a point on the flat line of table. There’s another point L/2 off to the side, to the right, say, that is the center of the square. P is then a distance x above that center point. The line of length a is from the wire to P, right? Now, what is B due to that segment of wire? It is perpendicular to this line, at P, do you see why? The magnetic field lines run in circles around the wire, so for any straight line running through the wire, the magnetic field at any point along that line is always perependicular to this hypothetical line. This applies to the line of length a running from the wire to P, too. However, because of all the other wires, the total field will be along the line of length x. So, focusing on B due to our segment, B^=r^ (just as it should, given the cross product), and we know that if we split B into components, one along x and one perpendicular, the perpendicular one will be canceled out by the rest of the square loop and the remaining one—the one that we will eventually quadruple to find Btotal—is the component along x.

Well, let’s complete the diagram, when there are so many things perpendicular to each other, you never know, things might come together where you’d never expect. Well, one triangle is formed by the lines of length a, L/2, and x, with linex⊥ lineL/2. Next, a line of length B comes off the linea- linex junction, perpendicular to a and away from x. On the same axis as x continues another line segment (parallel to x, you could say, though really the two segments are part of the same line) of length Bx. Now draw a line from the end of Bx to the end of B; that last line should be perpendicular to Bx, because it’s the other component. Look: now we have two triangles! And guess what? they’re similar triangles! If you don’t believe me, prove it yourself (could be good practice), keep in mind that right angles=90°, all angles in a triangle always add up to 180°, and if several angles added together form a straightedge, that means they add up to be 180°.

So anyway, the important thing is that if they are similar triangles, Bx/B=(L/2)/a => Bx=B·L/2(x²+L²/4)=μ0I L²/4π(x²+L²/4)(4x²+2L²). And finally, (bum-ba-bum!) Btotal=4Bx=0I L²/4π(x²+L²/4)(4x²+2L²).

1. A simple pendulum with a 1.5kg bob that has a charge of 1.0µC is deflected 0.20 radians in a horizontal uniform electric field. What is the magnitude of the electric field?

Solution: Show

2. A 1.0kg ball that has a charge of 0.50µC is dropped 50.cm (0.50m) through a horizontal uniform electric field and lands 5.0cm from directly below where it was dropped. What is the magnitude of the electric field?

Solution: Show

3. Two parallel metal plates in a vacuum are oppositely charged so that a uniform electric field of magnitude E exists between them. A neutral ping pong ball (radius R) falls into the thin space where there is the uniform electric field. What would you expect the electric flux through the ping pong ball’s surface to be?

Solution: Show

4. There is a dipole in a cavity in a (nearly) solid metal sphere that itself carries no charge. What is the net charge on the outside surface of the metal sphere? And the inside surface?

Solution: Show

5. A thick metal shell has a charge at it's center (magnitude Q) and a radius of R. A circle is drawn on it and a laser is used to slice off the protruding section in the circle, which is measured to have a (flat) radius r. If the flux through this flat circle is measured to be Φ, what was the flux through the larger area of the surface of the protruding section?

Solution: Show

6. A solid rod of charge with a volume charge density of ρ is miles long and has a radius of R. It has a small spherical bubble of (neutral) air radius r inside it, the center of which is d from the central axis of the rod. Determine the electric field on the outer edge of the bubble (that is, the point on the surface of the bubble closest to the outer edge of the rod), assuming the bubble is far from the ends of the rod.

Hint: Show

Solution: Show

7. A 2Q point charge with a mass of m is placed a distance D away from a Q point charge, which has twice the mass of the original 2Q charge. They start at rest on a infinite horizontal frictionless plane. Determine their speed t seconds later.

Solution: Show

8. Determine the electric field at P, which is a distance D directly below a vertical rod of charge that has a length L and a linear charge density of λ (you may want to find the electric potential first).

Hint: Show

Solution: Show

9. Determine the electric field at P, which is a distance D directly above a solid disk of charge that has a radius of R, a thickness of T, and a volume charge density of ρ (you may want to find the electric potential first). Note: this one’s not real hard, but it’s kinda long.

Hint: Show

Solution: Show

10. A simple parallel plate capacitor, within which there is a dielectric of a dielectric constant K, has an area A and separation d. The dielectric is pulled out in such a way that the area where there is a dielectric decreases linearly at a rate of R (so the area covered by the dielectric at time t is A-Rt). Ignoring any fringing effects of the electric field, find Ceq as a function of time.

Hint: Show

Solution: Show

11. It takes work to increase the separation between plates in a parallel plate capacitor, but work will be done on a plate of dielectric material if it is moved from outside the capacitor to inside. If a standalone capacitor has a voltage V, an area A, and a separation d, and putting the dielectric material inside would require expanding this to D (so it increases the separation by D-d), and the dielectric has a dielectric constant K, find the maximum D for which you could still put the dielectric into the capacitor without doing any work.

Solution: Show

12. Two capacitors are connected in a static situation, both with a voltage of V across them. One has a capacitance of C1 and the other C2. The dielectric constant of a material is K, and this material is pulled out from the capacitor with a capacitance of C1. What is the new voltage across the capacitors?

Solution: Show

13. You are in the Materials Science program at SOU, working on the recently discovered ID10s class of carbon nanotubes (stands for Iodine Doped formula 10s). You make a small mistake and create a whole new class of carbon nanotubes, which your professor dubs ID10t for how it was found (well thats what he says, could be for who found it). It’s conductive, and you measure that when a voltage of 1 V is applied to a 10cm wire 1mm thick, a current of 0.8 A passes through, whereas when a voltage of 4 V is applied to it, a current of 3 A passes through. Insofar, does it appear to be an ohmic conductor? If so, find it’s resistivity, if not, explain why.

Hint: Show

Solution: Show

14. A how much resistance R does a fuse need to have to blow when the rms current is I amps on a standard 120V ac line, if the fuse will melt when a power of P runs through it? Remember that the power to blow the fuse is the peak power, whereas the current through it is rms.

Solution: Show

15. Alec Trikcurrent has a two 12V batteries. One connected to two light bulbs in series, of resistance 3Ω and 6Ω, respectively, where the 3Ω light bulb is connected to the positive terminal. The other battery is connected to a 10Ω light bulb and a 2Ω light bulb in series, also, with the positive terminal connected to the 2Ω light bulb. He also connects a fifth bulb of resistance 1Ω between the two circuits, with one electrode of the bulb connected to a segment of wire between the two other bulbs on one circuit and the other electrode of the bulb connected to a segment on the other circuit, also between the two light bulbs. Whats the current through and voltage across this fifth light bulb? See this illustration.

Solution: Show

16. As a continuation of the last problem, now say Alec adds a grounding wire to the positive terminal on each battery. What is the voltage across and current through the fifth bulb now? Note: the system of five equations and five variables is actually quite simply solved. Also, see this illustration.

Solution: Show

17-20. Indy Uctor has the following supplies:

• Two 25cm long, 1.0mm thick iron wires (ρ=9.71·10-8Ω·m).
• An uncharged 3.0F capacitor
• A 6.0V battery with internal resistance 0.95Ω.
• A 10H inductor, resistance 0.35Ω.

17. He connects the battery to the capacitor with the wires for 5.0 seconds. How much charge is now on the capacitor?

Solution: Show

18. He then connects the capacitor to the inductor with the wires. How long should he leave them connected so that the charge on the capacitor decreases to 10C?

Hint: Show

Solution: Show

19. He then quickly connects the inductor to the battery again, in such a way that the battery pushes against the current in the inductor. Assuming that there was no change in current in the time it took for him to flip the switch and connect the inductor to the battery instead of the capacitor, how long will it take for the current to decrease to 0?

Hint: Show

Solution: Show

20. What would the answer to the last problem be if, at the beginning of this series (problem 17), the battery were attached to the capacitor for 6 seconds instead of 5 seconds?

Solution: Show

21. You are in a space ship approaching a spherical charged dust cloud, radius R0. From outside, the electric field behaves as if all the charge were concentrated at the cloud’s center, indicating that it has a spherically symmetrical charge distribution. As you travel within, you find that, until you reach a distance R1 from the center, the electric field E is linearly related to the distance from the center R, with a proportionality constant K1 (so that in the spherical shell between R=R0 and R=R1, but not the inner sphere beyond, E=K1R). Determine the charge enclosed within the spherical shell bounded by R1 and R0.

Solution: Show

22. As a continuation of the above problem, determine the volume charge density of dust in the outer shell between R=R0 and R=R1, where E=K1R (it's a constant, but what is it's value?).

Solution: Show

23. A spherical cavity is found at the center of the dust cloud, where there are no charges, of radius R2. The volume charge density in the inner shell between R1 and R2 is found to be linearly related to the distance from the center, with a proportionality constant K2 (so ρ=K2R). Suddenly, as you come out of the dust cloud again, a short circuit destroys much of the onboard electronics, to the extent of partially damaging the emergency jet. Only your trusty graphing calculator, which is not connected to the ship's computer systems, is untouched. Since the emergency jet has been damaged, it can only be fired once with a specified strength. Given that just before the short circuit the field was E, and because you were making detailed scientific measurements, the ship wasn't moving very fast and the field at the moment is essentially the same, determine how far you are from the center.

Solution: Show

24. A particle of mass m, positive charge q, and inital velocity v0 to the right is in a uniform electric field E down and uniform magnetic field B out. Determine the (multi-equation) relationship between the magnitude of the velocity v and the time t (where ‘initial’ refers to when t=0). Note: it’s several differential equations, so if that’s what you got, don’t immediately assume you made a horrible mistake.

Solution: Show

25. A toroidal solenoid (a solenoid bent into a ring shape) with N turns lies flat on a table, the ends of the wires connected to a galvanometer. The radius of the toroid is R and the radius of the solenoid is r (so a solenoid, radius r, is bent into a circle of radius R). There is a uniform electric field coming directly out of the table, increasing at a rate of et. Assuming r is small compared to R, and N is quite large, what magnitude of voltage does the galvanometer read, as a function of time? (Yes thats right: E causes B causes V. I wonder if Dennis will have something like that on the test. No guarantee, but it's just the type of clever thing).

Solution: Show

26. You are Micro, the Human Molecule. As an EM wave passes by, an electron (initially at rest) on your molecular table jumps up then down, and an electron sliding by in front of you spins clockwise then counterclockwise. Which way was the EM wave going, to the left or the right? (Yeah, I know, kinda cheesy.)

Solution: Show

27. How great would the maximum electric field vector be of the light required to keep a 1.0g, 1.0m² 90% reflective piece of mithril floating above the ground? (g=9.8m/s²)

Solution: Show

28. A magnet of mass m is held a height h above a circular loop of wire, which has a radius of r and a resistance of R. The average magnetic field going down into and through the loop is B. If the magnet is then dropped, what is it’s velocity when it has fallen to h below the loop? Assuming that the magnet is perfectly symmetrical, so that when the magnet has fallen to h below the loop, the average magnetic field is B going up through the loop.

Solution: Show

29. A five turn coil (resistance 10.Omega;, radius 5.0cm, thickness 1.0cm) connected to a transformer is on the table, with a metal band fit snugly around it. The metal band has a tiny gap, that requires 500V across it to spark. What ratio of turns is required for the tranformer to be able to turn standard 120V, 60Hz ac current into whatever is required to spark the gap in the metal band? Assume the ac current varies in a "sharktooth" patter rather than sinusoidally, so that dV/dt for standard ac current is 2·120V/(1/60Hz)=2·120V·60Hz.