How Does a Positive Charge Affect Magnetic Field Components?

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http://img220.imageshack.us/img220/5343/picture65653vt9.jpg









It says on top a particle with postive Charge q=+3UC. Thanks

All right, i know how to get all the answers which to the other stuff, i just need to know whether or not 7-9 is correct, my reasoning is X will increase since it is postive charge and Y will decrease, If it was Negative then it would be the Opposite. I don't really know about the Z componet and for Number 5 i think it should be coming out of the page. Thanks

 

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So is there a formula for thi stuff to figure?

 

[berkeman]

So is there a formula for thi stuff to figure?

Yes, and as far as I can see, the only thing you got right is that the Lorentz force on the positively charged particle is vertical at you out of the page.

Does your text cover the Lorentz force? What is the vector equation for the Lorentz force? If you don't have it in your study notes or text, maybe try wikipedia first, and then google. The Lorentz force pops up in a lot of E&M work.

 

[Hargoth]

The force which acts on a moving charge in a magnetic field is perpendicular on both the field and the velocity. So there is, for example, no force and no acceleration in the direction of the field and v_y is constant. To figure it all out you can look at the Lorentz Force:
F = q [\vec v \times \vec B ], which this particle experiences in the field.

 

[Alt+F4]

The force which acts on a moving charge in a magnetic field is perpendicular on both the field and the velocity. So there is, for example, no force and no acceleration in the direction of the field and v_y is constant. To figure it all out you can look at the Lorentz Force:
F = q [\vec v \times \vec B ], which this particle experiences in the field.

We have no book, Ok so X Would Decrease, Y would Increase and Z would be constant
since according tho this http://en.wikipedia.org/wiki/Image:Lorentz_force.svg the postive charge will go up the field thus have Y increase, and X decrease. Correct?
So F= (250*1.8)*3 = 1350 what does that number tell me

 

[Hargoth]

Correct.

 

[Alt+F4]

Correct.

Can u see if my explanation actually makes sense? Thanks a lot i just added something in

 

[Hargoth]

Nope, because on the picture the magnetic field comes out of the "paper". Your first answer was correct. Your force isn't correct either.
Just calculate the Lorentz-Force with the formula above.

 

[Hargoth]

Ok so i assume u mean Calculate Magnitude like question 5 well for that one

I used F=QvBSin Thetha = )3*10^-6)(250)(1.8)(Sin122) = .00114N


So X would have to Decrease, Y Increase and Z remain Constant?

No. Y is perpendicular to the field, remember? Do it with the formula above, you get the magnitude and even the vecor, with immediately shows you in which directions a force exists.

 

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Okay I got it, X-Decrease, Y COnstant, Z Increasing

 

[Hargoth]

Sorry, I meant Y is in the direction of the field, of course. You already said that the force points out of the paper. From your picture I'd say that's the z-direction.

Let's calculate the force on the particle. y is the direction of the field, so

\vec{B} = B \vec{e_y}.

The velocity is \vec v= \cos(32)v \vec{e_x}+\sin(32) v \vec{e_y}.

So the Lorentz-Force is \vec F = q[\vec v \times \vec B] = q[(\cos(32)v \vec{e_x}+\sin(32) v \vec{e_y}) \times B \vec{e_y}]=qBv cos(32) \vec{e_z}
What does this tell you?

 

[Alt+F4]

Sorry, I meant Y is in the direction of the field, of course. You already said that the force points out of the paper. From your picture I'd say that's the z-direction.

Let's calculate the force on the particle. y is the direction of the field, so

\vec{B} = B \vec{e_y}.

The velocity is \vec v= \cos(32)v \vec{e_x}+\sin(32) v \vec{e_y}.

So the Lorentz-Force is \vec F = q[\vec v \times \vec B] = q[(\cos(32)v \vec{e_x}+\sin(32) v \vec{e_y}) \times B \vec{e_y}]=qBv cos(32) \vec{e_z}
What does this tell you?

this whole X-Decrease, Y COnstant, Z Increasing

 

[Hargoth]

Why should the x-velocity decrease? Is there a force?

 

[Alt+F4]

Why should the x-velocity decrease? Is there a force?

No no force at all, but doesn't the postive 3 UC charge have an effect?

So X and Y would be Constant and Z increasing


I guess since The speed of the particle is constant then no force is acting on it.

 

[Hargoth]

No no force at all, but doesn't the postive 3 UC charge have an effect?

No. Newton's law: If there's no force, there's no change of velocity. In our case, there is no force in the x- and y-direction, only in the z-direction, so what does this mean?

 

[Alt+F4]

No. Newton's law: If there's no force, there's no change of velocity. In our case, there is no force in the x- and y-direction, only in the z-direction, so what does this mean?

Constant X and Y, ok i tell u what i have another quiz from a previous semster where the Arrow is pointing up like open the arrow 122 degrees more and her answers are X decreases, Y increases and Z is constant. so that is why there was confusion