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GR8677 #29
Problem
 GREPhysics.NET Official Solution Alternate Solutions

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.)

Alternate Solutions
 mahdisadjadi2012-09-27 10:06:49 The answer is definitely B. As you can see, test charge is negative and the official solution points it out. But in the answer this is not considered. Assume that: $\mathbf{I}= I\hat{z}$ In this situation, direction of magnetic field is in the right hand side along $-\hat{x}$ and is along $\hat{x}$ in the left hand side of wire, so we can write , by Loentz force law(we take q to be positive and enter a minus to indicate negativity of charge): $\mathbf{F}=-q\vec{v}\times\vec{B} = - qB (v_{y}\hat{x}+v_{x}\hat{y})$ Force should be parallel to the current, so we take $v_{x}=0$ and $\mathbf{F}= F \hat{z}$, where $F$ should be positive. So, $v_{y}$ is negative and charge is moving away from the wire. Reply to this comment
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kronotsky
2018-10-23 03:44:30
Right hand rule trick: if you put your thumb in the direction of I, and curl your fingers up into a fist, the fingers point along B. We want $q \\vec{v} \\times \\vec{B} = \\vec{I}$. Cyclic permutations and setting $q = -1$ gives $\\vec{I} \\times \\vec{B} = \\vec{v}$.
QuantumCat
2014-09-02 14:01:17
Consider $I$ = $I_0{\hat k}$ so that $B$ = $B_0 {\hat \varphi}$ . Because q goes to -q, consider that $- \hat k$ = $\hat{v} \times \hat{\varphi}$ so $\vec{v} \rightarrow v_0 (-\hat{r})$ leading to choice A
2012-09-27 10:06:49
The answer is definitely B. As you can see, test charge is negative and the official solution points it out. But in the answer this is not considered.
Assume that:
$\mathbf{I}= I\hat{z}$
In this situation, direction of magnetic field is in the right hand side along $-\hat{x}$ and is along $\hat{x}$ in the left hand side of wire, so we can write , by Loentz force law(we take q to be positive and enter a minus to indicate negativity of charge):
$\mathbf{F}=-q\vec{v}\times\vec{B} = - qB (v_{y}\hat{x}+v_{x}\hat{y})$
Force should be parallel to the current, so we take $v_{x}=0$ and $\mathbf{F}= F \hat{z}$, where $F$ should be positive.
So, $v_{y}$ is negative and charge is moving away from the wire.

 walczyk2012-10-13 13:06:27 Uh if Vy is negative then it is moving TOWARD the wire, not away from it. Also the answer is A, if ETS gave the wrong answer you wouldn't be the only one discovering this. This test is very very old.
 eris12015-10-08 14:48:11 Define the cable as laying on the $z$-axis at the origin of the $y$ and $x$ axis. Imagine the particle is moving strictly along the y axis, starting at a point $p_y$>0. With a negative velocity (in the negative $y$ direction), the particle is moving towards the wire and the $B$-field it encounters is in the positive $x$-direction. If the particle starts at $p_y$<0 then the direction of the $B$-field it encounters is in the negative $x$ direction (canceling the \"-\" from the charge), thus following that the velocity of the particle must be positive (for a positive force), moving again towards the wire.
 camarasi2017-10-25 14:05:56 The x-component of the B-field as you\'ve taken it is negative, so you\'re off by a sign.
Fily
2011-04-07 04:12:37
This is more simple just z=-?*Phi(The angle).So its ultimately give ?=-r
shafatmubin
2009-10-31 18:12:46
Fleming's left-hand rule (thuMb for motion, First for field, seCond for current) is used to find direction of motion (i.e. force applied) when current and field directions are known.

So the LHR will work here, if one takes into account of the direction of the CURRENT produced by negative charges (i.e. opposite to velocity direction).
dean
2008-10-09 21:45:23
I may be mistaken, but it seems to me the LHR yields the wrong answer here if used consistently (i.e. twice). Better to stick with the right hand (sorry southpaws) and remember sign.
 neon372008-11-02 15:41:09 not really you are probably trying to figure out the direction of the magnetic field also with LHR. The direction of the field is always with the RHR. You could find the direction the test charge should go given the direction of the field and current, with LHR. I would also suggest sticking to RHR on the real exam. Might be confusing.
darox
2006-11-29 06:59:15
actually, it cannot be B. the direction of the particle would be not parallel, but anti-parallel.
 zaharakis2007-01-05 09:31:42 Anti Parallel is still parallel
 FortranMan2008-10-19 13:14:30 I thought anti-parallel is still parallel too, until I looked at options (A) and (B). This problem is deceptively simple, people. Remember that if the charge is moving towards the wire its velocity vector would be negative, but this negative can be canceled out if you are using a negative test charge, resulting in the positive force direction parallel to the positive current direction. In short, the sign of the charge can affect the vector.
zaharakis
2006-11-02 14:25:53
The answer could also be B. The question states parallel to the direction of the current. B would produce a force in the opposite direction to the current but this is still parallel.
 zaharakis2007-01-05 09:31:14 Anti parallel is still parallel.
 tau17772008-10-31 19:30:03 yeah, i did not realize this the first time i took the test. and as i was redoing it today it hit me, this confusing thing about parallel. i feel that they should be more direct but i guess that's just the ETS. i'll have to keep an eye on my common sense of things and try to see it their way, least until test day. bottom line: to the ETS antiparallel and parallel are different.
 gt20092009-06-22 06:04:21 Anti-parallel and parallel are different.

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