# What is the magnetic field generated by these two particle beams?

A uniform beam of positively charged particles is moving with a constant velocity parallel to another beam of negatively charged particles moving with the same velocity but in opposite direction separated by a distance d. Then, how should be the variation of magnetic field B along a perpendicular line drawn between the two beams?(For better view, imagine positive beam to be x-axis and negative beam to be y=1 line.)

I know that electric field will be there. But how the magnetic field will be generated and how will it vary?

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BvU
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Hello Anurag, $\qquad$ $\qquad$ !

Please post in homework and use the template -- it's mandatory

What do you know of the B-field generated by a current $i$ along the x-axis ?

Hello Anurag, $\qquad$ $\qquad$ !

Please post in homework and use the template -- it's mandatory

What do you know of the B-field generated by a current $i$ along the x-axis ?
Thanks. I will take care of this from now on.

Hello Anurag, $\qquad$ $\qquad$ !

Please post in homework and use the template -- it's mandatory

What do you know of the B-field generated by a current $i$ along the x-axis ?
It follows Biot-Savarts Law. That is ${μ(i×dl)} /{r^2}$. (I am not able to produce it in fraction expression)

anorlunda
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I moved it to homework, but you must still show your effort before our helpers are allowed to help.

I moved it to homework, but you must still show your effort before our helpers are allowed to help.
But I only know about current producing Magnetic field and not about charge beam producing magnetic field

Hello Anurag, $\qquad$ $\qquad$ !

Please post in homework and use the template -- it's mandatory

What do you know of the B-field generated by a current $i$ along the x-axis ?
But it is charge beam and not a current carrying wire. Are both the above to be the same situations for magnetic field? If yes, then how?

BvU
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See #5. Remember that current = moving charge

See #5. Remember that current = moving charge
So can we solve it using Ampere's circuital law?
As $\vec{B}.{d\vec {l}} =μi=\frac{qv} {x}$ But if it is so, then what should we take x?

BvU
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But it is charge beam and not a current carrying wire
What's the difference ?

Let a to b be on the x-axis, one unit of distance apart (1m).
How much charge per unit time goes from a to b on the x-axis if there is a current $i$ ?
And how much charge per unit time goes from a to b on the x-axis if there is a beam of particles with charge q that move with velocity $v$.

Conclusion ?

BvU
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$\vec{B}.{d\vec {l}} =μi \$ I recognize. $μ i=\frac{qv} {x}$ misses a $\mu$ .
What is the direction of $\vec v$ in your exercise ?
So what direction for $x$ ? for $\vec B$ ?

What's the difference ?

Let a to b be on the x-axis, one unit of distance apart (1m).
How much charge per unit time goes from a to b on the x-axis if there is a current $i$ ?
And how much charge per unit time goes from a to b on the x-axis if there is a beam of particles with charge q that move with velocity $v$.

Conclusion ?

Okay. I understood. From what I am able think is that the field varies radially from each current and that we can use principle of superposition?

BvU
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You got it.
But be careful... $\vec{B}.{d\vec {l}} =μi$ I recognize. $\μ i=\frac{qv} {x}$ misses a $\mu$ .
What is the direction of $\vec v$ in your exercise ?
So what direction for $x$ ? for $\vec B$ ?
My mistake for μ. Please see # 12.

You got it.
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