Find the current induced in a wire loop by a nearby current

[Helly123]

Homework Statement

Homework Equations

The Attempt at a Solution

When the circuit moves creates magnetic force that moves the some charges in the wire and that is a current.
That is what i thought.

F = Q.v B sin theta
B = ( permeability Of vacuum* I)/(2pi.r)
But how to find the I?

Thank you very much

 

[BvU]

Please type your problem statement, list of all varables, knowns/unknowns and relevant equations. The picture is illegible.

 

[BvU]

But how to find the I?

You don't have to. All the answers are given in terms of $I$, not as numbers.

 

[Helly123]

Please type your prob;lem statement, list of all varables, knowns/unknowns and relevant equations. The picture is illegible.

I am sorry. I should have checked it. now, I have replaced the picture. please have a look

 

[BvU]

Same picture, same sharpness, though...

 

[Helly123]

You don't have to. All the answers are given in terms of $I$, not as numbers.

ok, I = V/R
the Resistance = R

is that anything to do with B?

 

[Helly123]

there is circuit ABCD shaped as a square next to a wire
circuit description :
with AB (vertical) = 2a
BC (horizontal) = 2b
the center is O.
O to B is b ( B to C = 2b)

what is the current flow in ABCD circuit, when the circuit reached r away from wire to its center O?
(the circuit move with speed v to right, AB still parallel to wire)

the question says B (Magnetic flux density) to be negligible

 

[BvU]

You should distinguish the $I$ in the long wire from the currrent in the ABCD loop. Still can't read the picture...[edit] better now...

 

[BvU]

And you need a relevant equation for V, the emf induced in the loop

 

[Helly123]

V = emf = electromotive force = L.B.v
emf = L. $\frac{µ0 * I}{2 \pi r}$.v

emf = 2a. v. $\frac{µ0 * I}{2 \pi (r - b)}$​

I ind = Emf/R = av $\frac{µ0 * I}{\pi R (r - b)}$

 

[BvU]

And where is that emf ? In AB, BC, CD or DA ?

 

[Helly123]

And where is that emf ? In AB, BC, CD or DA ?

it is on AB

 

[BvU]

What about the others ?

 

[Helly123]

Can you please elaborate? Because i don't understand

 

[BvU]

You found an emf on AB. Does any of the other sides of the square also have an emf ?

 

[Helly123]

I don't really understand the question, lol. Is it a circuit moves from a wire in magnetic field? If yes, there is B. By right hand rule the B goes into the page. Magnetic force goes to the left, as velocity goes to right.

 

[Helly123]

Usually it is a wire moving in magnetic field and induce a current. That is why i don't get this question..

 

[BvU]

There really are four connected wires in a square loop. You looked at only one and stopped looking !

 

[Helly123]

Yes. So in emf = Blv sin theta
I think the theta refers to angle between movement and magnetic field.. so, for BC and AD what is the theta? And which direction is magnetic field ..

 

[BvU]

And which direction is magnetic field ..

Keep going...

And check here

 

[Helly123]

Emf BC = bv$\frac{µ0.I}{\pi (R-b)}$
Emf CD = av$\frac{µ0.I}{\pi (R+b)}$
Emf AD = bv$\frac{µ0.I}{\pi (R-b)}$

 

[BvU]

No sir. Only one out of three is (almost ) correct .

You want to look carefully at the here link in #20.

And in fact, also at the keep going one...

 

[Helly123]

The only one right is CD and the answer is B? But i don't understand why. Why there is no voltage in BC and AD?

 

[BvU]

BC and DA, that is. They correspond to the middle picture where no emf is induced along the wire.
You're very quick to pick the B answer and not A, but I suppose you worked it out properly ?
(did you understand my 'almost' in #22 ?)

The way I remember things: I just start with the Lorentz force

To keep a long story short I add over a Thousand words:View attachment 168089
Big picture, big importance (imho)

Both pictures:

• Positive $x$ direction is to the right
• Magnetic field is down. Consider that the positive $y$ direction
• Big arrowheads: in the $x\, y$ plane
• Smaller arrowheads in the $\hat z = \hat x \times \hat y$ direction (a very simple cross product)
• The coordinate system is right-handed
• The Lorentz force is $\vec F = q\, \left (\vec E + \vec v \times \vec B \right )$
• There is no E field
• Right hand rule: in $\vec a \times \vec b = \vec c$
  
  a thumb
  b index
  c middle -- points in the direction of c​
• Corkscrew rule: in $\vec a \times \vec b = \vec c$
  
  turn $\vec a$ over the smallest angle towards $\vec b$ : Corkscrew goes in the direction of c​

Top picture:

• We see a wire carrying current to the right. Current is moving charge. Moving charge has a "velocity".
• We consider mobile positive charges in the wire
• The Lorentz force is $\vec F = q\, \vec v \times \vec B$
• All moving positive charges in the wire feel this Lorentz force
• Together they "push" in the direction of $\vec v \times \vec B$
• mutatis mutandis for moving negative charges: they move in the negative x-direction and the resulting Lorentz force points in exactly the same direction (- q x -v = +q x +v)

Bottom picture:

• We see a wire being dragged in the positive $z$ direction with a velocity $\vec v$.
• We consider mobile positive charges in the wire
• The Lorentz force is $\vec F = q\, \vec v \times \vec B$
• All moving positive charges in the wire feel this Lorentz force
• Together they "push" in the direction of $\vec v \times \vec B$, which is the negative $x$ direction
• This "push" is an electromagnetic potential (emf) that can cause a current.

 

[Helly123]

Such a new thing for me to know.. thank you so much. So, the velocity direction is opposed to force?
Why only positive charges feel Lorenzt force?
Is the moving of charges produce Lorentz force or the positive charges moves to fight Lorentz force?

 

[Helly123]

You should distinguish the $I$ in the long wire from the currrent in the ABCD loop. Still can't read the picture...[edit] better now...

What is $I$ in the wire refers to? To current in straight wire? And we look for current in square circuit?

 

[Helly123]

BC and DA have force but no voltage since it moves parallel to v. AB and CD moves perpendicular to v. And have voltage no force.

 

[BvU]

#22:
The 'almost' was because of the sign. It is opposite to that induced in AB.

#25:

So, the velocity direction is opposed to force?

No. Velocity $\vec v$ and magnetic field $\vec B$, are both perpendicular to the resulting force $F$.

Why only positive charges feel Lorentz force?

Not what I meant. What I meant is: one doesn't want to confuse oneself unnecessarily by looking at negatively charged moving charge carriers. So if a current $\vec I$ is given, work with positive charge carriers moving in that direction.

The Lorentz force $\ \vec F = q(\vec E + \vec v\times \vec B)\$ works on ALL charges, negative (q<0) or positive (q>0)
( also on neutral charges (q=0) )

Is the moving of charges produce Lorentz force or the positive charges moves to fight Lorentz force?

Motion of charge produces a magnetic field.
Positive charges move under the influence of a Lorentz force -- if they can. In a conductor (a wire), the positive charge carriers generally are not free to move. Electrons, the negative charge carriers, can move and thereby generate an emf.

I think I understand what you are trying to say here (correct me if I miss): Lorentz force on CD pushes electrons down and positive ions up. Positive ions can't move, electrons can. They don't move without limits (they don't pile up because they repel each other, and there is a resistance)​

#26: The exercise in post 1 gives a current $I$ in the long wire and asks for the current in loop ABCD; with resistance $R$ given, that is equivalent to asking for the emf divided by $R$.

#27:

BC and DA have force

They do not. Their net charge is 0.

AB and CD move perpendicular to v

No. Their direction of motion is the direction of $\vec v$. Their orientation is perpendicular to $\vec V$.

 

[Helly123]

What makes the sign different? Since, B, v still the same?
And where the 'b' comes from the options?

 

[Helly123]

What makes the wire have magnetic force?
Why the question said that the magnetic flux density is negligible?