<p>you know that I1=2I2 because since you are told there is no current through the ammeter, you can imagine it as two parallel wires.</p>
<p>@ JDong217
For problem in post #191, remember that when your inside the earth, the M in GMm/r^2 isn’t constant (you have to find the ratio of M that is actually inside r). So to find the kinetic energy, which is the work since it starts at rest and change in KE=work, the integral is w=-integral of (GmM*(r^3/R^3)/r^2 where R is the radius of the earth and r is the current distance from the center. When you integrate from R to 0, you get GmM/2R</p>
<p>and II is equivalent to III when you substitute that g=GmM/R^2</p>
<p>@BuddyMcAwesome</p>
<p>Doesn’t the current drawn through each wire depend on the total resistance of the wire? So how would we know how much current is drawn through each wire if we don’t know the total resistance of the bottom wire?</p>
<p>
</p>
<p>Yes I know that, but why can’t E be an answer?</p>
<p>the idea is that the current draw through each side of each wire must be the same.</p>
<p>to clarify Buddy’s explanation of the circuit problem, because the ammeter reads 0, you know that there is no potential difference between the two ends (or else this would drive movement of charge). Because you know the two ends connecting to the ammeter are equal (not necissarily 0), you can write that the voltage across the 3 ohm resistor is equal to the voltage across the R resistor, and the voltage across the 1 ohm resistor equals the voltage across the 2 ohm resistor. Also you know that that current through the 1 and 3 resistor is the same, and the current through 2 and R is the same. So you get 3(I1)=R(I2) and (I1)3=(I2)R… and then solve</p>
<p>omg im so glad im not taking E&M…</p>
<p>I just started self-studying E&M 2.5 weeks ago… I need a 5 to get any advancement credit :/</p>
<p>@ math, for the second problem, it can’t be E because the work is non-conservative, therefore you cannot say that the change in kinetic energy is related to the change in potential energy.</p>
<p>
</p>
<p>It looks like aznjunior did a good explanation of why III is true. To find if II is true, just plug in “g” into III. We know g=GM/R^2, so (GmM)/(2R) = mgR/2</p>
<p>@BuddyMcAwesome:
How is the current drawn through each branch the same if you’re saying that I<em>1=2I</em>2?</p>
<p>the current through the top circuit (through the 1 and3 ohm resistor) are the same. The current through the bottom circuit ( R and 2 ohm) are the same. However the current through the top does not equal the current through the bottom… The voltage of the top circuit equals the voltage of the bottom.</p>
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<a href=“http://i40.■■■■■■■.com/312cmc1.jpg[/IMG]”>http://i40.■■■■■■■.com/312cmc1.jpg
![http://i40.■■■■■■■.com/312cmc1.jpg[/IMG]](http://i40.tinypic.com/312cmc1.jpg%5B/IMG%5D){kind=link}
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<p>A rigid rectangular wire loop ABCD carrying current I1 lies in the plane of the page above a very long wire carrying current I2, as shown above. The net force on the loop is:
(a) toward the wire
(b) away from the wire
(c) Toward the left
(d) toard the right
(e) zero</p>
<p>I know it’s a according to the answer key buy why? I know parallel currents attract so the segment AD has force toward the wire, but then wouldn’t the force on segment BC negate the force on AD because it’s in the opposite direction?</p>
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<a href=“http://i42.■■■■■■■.com/2cekbwj.jpg[/IMG]”>http://i42.■■■■■■■.com/2cekbwj.jpg
![http://i42.■■■■■■■.com/2cekbwj.jpg[/IMG]](http://i42.tinypic.com/2cekbwj.jpg%5B/IMG%5D){kind=link}
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<p>A bar magnet and a wire loop carrying current I are arranged as shown above. In which direction, if any, is the force on the current loop due to the magnet?</p>
<p>Can someone please explain to me how you would apply Lenz’s law to this type of problem?</p>
<p>@JDong217
For the first one, the current at the bottom creates a magnetic field into the page at the loop. To find the force on the loop by this magnetic field on the current, use the right hand rule. Since current is going to left, and magnetic field is into the page, force is going to the bottom of the page. This force is greater than the force on the top part of the loop which points up because the bottom segment of the loop is closer to the current creating the magnetic field.</p>
<p>Predictions for E+M</p>
<p>FRQ
1: Electrostatics(most likely capacitors)
2: Some sort of RC, R, or RL circuit
3. Induction/Inductors/Faraday’s law problem</p>
<p>MC:</p>
<p>Lots of Electric Feild and Circuit type problems. Not very much magnetism.</p>
<ol>
<li>The magnetic field due to a current carrying wire is not constant, so the force is greater for the segment that is closest to the wire. The field, and therefore the force, falls off 1/r for a constant current. as you move away from the wire.</li>
<li>Lenz’s law is for determining induction, it doesn’t apply here. Just use the right hand rule to determine that the north pole of the loop is towards the magnet. Therefore they repel.</li>
</ol>
<p>Maximum displacement is at the amplitudes/where velocity is minimum and the object is about to change directions.</p>
<p>Minimum potential energy is when KE is max, which is when the particle is moving the fastest, which is at equilibrium when x=0 and the acceleration is a minimum</p>
<p>You know how for mechanics they always have problems where it has a spring and then points x=-a, x=0 and x=a … can someone explain what is going on at each point?</p>
<p>like where is maximum displacement, minimum potential energy of the spring, potential energy of the block</p>
<p>for some reason those always get me</p>
<p>x=a/-a: max/min displacement min/max acceleration (because a = -w^2*x)</p>
<p>x=0: max/min velocity (because 0 potential energy in the spring at x=0, so speed is fastest there since all energy is in the form of kinetic energy)</p>
<p>EDIT: Nevermind I got it…</p>