# Correct statement regarding a conductor in a magnetic field

songoku
Homework Statement:
A stationary metal conductor is carrying a current
in the direction to the right in the plane of the
paper as shown. It is in a region of uniform
magnetic field pointing perpendicularly into the
plane of the page. Which of the following
statements is correct?
a. P has a lower potential than Q
b. P and Q are at the same potential
c. P is at a positive potential
d. P has a higher potential than Q

(diagram given below)
Relevant Equations:
Fleming's Left Hand Rule By using Fleming's Left Hand Rule, I got the force acting on proton is directed upwards so my answer is (d) but the answer key is (a). So the force acting on proton is actually downwards?

Thanks

Homework Helper
I got the force acting on proton is directed upwards
That is correct. The Lorentz force acts in the upward direction.

To speak of this in terms of potential is confusing and didactically unwise. It is unclear what the exercise composer means: book answer (a) hints at a fictitious potential acting on the moving charge carriers. But there is only the Lorentz force.

One can equally well claim that (b) is correct: in an ideal conductor the potential is the same everywhere.

It is not even clear where P and Q are located exactly: outside the conductor ? In that case your arguments are valid IMHO

Perhaps the book answer is simply wrong ?

• songoku
songoku
That is correct. The Lorentz force acts in the upward direction.

To speak of this in terms of potential is confusing and didactically unwise. It is unclear what the exercise composer means: book answer (a) hints at a fictitious potential acting on the moving charge carriers. But there is only the Lorentz force.

One can equally well claim that (b) is correct: in an ideal conductor the potential is the same everywhere.

It is not even clear where P and Q are located exactly: outside the conductor ? In that case your arguments are valid IMHO

Perhaps the book answer is simply wrong ?

The explanation of the book: By Fleming’s Left Hand Rule, electrons will be deflected upwards

Homework Helper
https://www.electrical4u.com/fleming-left-hand-rule-and-fleming-right-hand-rule/ says something about the force on the conductor, not necessarily about the force on the electrons (though I think in the end they are one and the same).
I find it confusing to use anything but right-hand rules.

The Lorentz force is ##\vec F = q(\vec E + \vec v\times\vec B)## and with current to the right, ##\vec v## is to the right for positive charge carriers, so ## \vec v\times\vec B## is upward.
If the protons (better to speak of positive charge carriers) would move, the force would be upward.

In a metal conductor, conductance is not through moving protons, but through moving electrons. Current to the right means electrons drifting to the left, so ## \vec v\times\vec B## is downward, but ## q\vec v\times\vec B## is upward.

Your exercise composer finds this upward force on the electrons causes a potential difference in the conductor. With negative charge buildup at the upper side P would be negative wrt Q.

• songoku and berkeman
Gold Member
I believe the answer must be (b).

Yes, the current will be deflected upwards. And thus there will be more positive charge near the upper side of the conductor, and more negative charge on the lower side.

But it's still a conductor. Conductors cannot support a potential difference without current flow. So once the system has had a tiny bit of time to stabilize, the potential difference should be zero.

To put it another way, imagine we start the system with the wire carrying current with no external magnetic field. There would be no potential difference, clearly. Then we turn on the external magnetic field: a potential difference is generated by the Lorentz force which causes current to move upward, meaning that the top (P) is at a lower potential than the bottom (Q).

But the charge flows in response to that potential difference. It keeps flowing towards P until the potential difference is eliminated. Once the system stabilizes so that no more current is flowing towards/away from P/Q, it has to be at zero potential difference across the conductor.

• songoku
Mentor
I think the problem here is the definition of the electric potential. There will be a higher electron density at the top, leading to an electric field in the conductor - balancing the force from the magnetic field. Does that count as potential difference, if the electrons (on average) feel no force in this direction?

If the magnetic field is limited to the area of the conductor and you connect the opposite sides outside the magnetic field you get a current flow. Going by that definition there is a potential difference and (a) is the correct answer.

• scottdave and songoku
songoku
but ## q\vec v\times\vec B## is upward.
Sorry I don't understand this part. I always think that if force on positive charge carriers is upwards, then force on negative charge carriers will be in opposite direction (downward)

There will be a higher electron density at the top
Why is electron density higher at the top? Using fleming's left hand rule, the force acting on negative charge carriers will be downward so electrons gather at the bottom part?

Thanks

Homework Helper
Gold Member
• songoku
Homework Helper
Sorry I don't understand this part. I always think that if force on positive charge carriers is upwards, then force on negative charge carriers will be in opposite direction (downward)
The 'convention' for direction of current is the direction that positive charge carriers would move. So negative charge carriers move in the direction opposite to the direction of 'the current'.
In other words: electric current to the right means electrons drifting to the left.

Fleming's rule works for 'the current' in the sense of moving positive charge carriers.

Why is electron density higher at the top? Using fleming's left hand rule, the force acting on negative charge carriers will be downward so electrons gather at the bottom part?
Again, I advise to forget any and all hand rules, especially the left hand ones and work with vector products. A way to avoid hand rules (with their associated problem: which finger is assigned to what) is to think of the corkscrew: for ##\bf a \times b ## turn ##\bf a## over the smallest angle in the direction of ##\bf b## and the vector product points in the direction a corkscrew goes.
Admittedly the corkscrew rule is a right hand rule​
And to make things worse they invented left-hand corkscrews for left handeds, but fortunately they are extremely rare (those corkscrews, that is ).​

As Rudy (I find @rude man a contradiction ) so correctly posts: check out the Hall effect.

• • scottdave, vela and songoku
songoku
I think I get it. Thank you very much for all the help

• BvU
Homework Helper
Gold Member
Again, I advise to forget any and all hand rules, especially the left hand ones and work with vector products.
Agreed, in fact I usually forget about hands altogether. I set up a coordinate system and use the well-worn determinant to evaluate curl. For some reaon I have no problem figuring out the i x j = k, -i x j = -k etc. etc. sequences.
(I find @rude man a contradiction )
Thx - but where there's smoke ..." : Last edited:
• BvU
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