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This is a question I've in my textbook

Imagine that you are sitting in a chamber with your back to one wall. An electron beam, moving horizontally from back wall towards the front wall, is deflected by a strong magnetic field to your right side. What is the direction of magnetic field?

As electron is moving backward to forward direction, charge would be moving forward to backward. Force is applied on right side and thus, by using fleming's left hand rule I can find the direction of magnetic field to be downwards. Here is the picture picture depicting the movement of an electron in a magnetic field

But here, I'm confused. As electron is moving to the right, current must be moving to the left. For other questions which I've solved, force was always applied to current. So as current is moving forwards to backward and it deflects to left side, then by using fleming's left hand rule the magnetic field should be upwards. Why then it is downwards. I'm probably misunderstanding how force is applied to the charge in a magnetic field..

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2 Answers 2

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(1) You are applying Fleming's LH rule correctly in your paragraph immediately above the diagrams.

(2) In your paragraph below the diagrams, I suspect that you're being misled by your use of the terms 'left and 'right' as when you say "So as current is moving forwards to backward and it deflects to left side..". It's the directions shown on the diagram that matter: point your second finger (left hand) in the current direction and your thumb in the deflection direction, then the first finger will point into the page.

(3) John Ambrose Fleming was an engineer and his 'hand rules' were intended for use by engineers dealing with currents and emfs in wires in motors and generators. Physicists view the force on a current-carrying wire in a magnetic field as arising from the force on each moving free electron in the wire. This force is a special case of the so-called magnetic Lorentz force on a charge moving in a magnetic field, as long as the field has a component at right angles to the charge's velocity.

A particle of charge $q$ moving with velocity $\vec v$ in a magnetic field of flux density $\vec B$ experiences a magnetic Lorentz force, $\vec F$, given by $$\vec F=q\ \vec v \times \vec B.$$ This equation, written in vector algebra notation, gives both the magnitude and direction of the force. For a particle outside a wire it is more convenient and more appropriate than Fleming's Left Hand Rule.

[$\vec v \times \vec B$ is a so-called 'vector product'. The result of performing the $\times$ operation is another vector, whose direction is given by this rule... Represent $\vec v$ and $\vec B$ by arrows with their tails at the same point, O. Call the angle between them that is less than 180° '$\theta$'. Rotate the first vector, $\vec v$, through $\theta$ about O so it now co-incides in direction with the second vector, $\vec B$. The direction of $\vec v \times \vec B$ is that in which an ordinary bolt with a right-handed thread would advance or retreat through a threaded hole if the $\vec v$ arrow were a spanner turning the bolt. The magnitude, $|\vec v \times \vec B|$, is $vB\ \sin \theta$. Multiplying $\vec v \times \vec B$ by a negative scalar, $q=-|q|$, reverses its direction – as well as multiplying its magnitude by $|q|$.]

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  • $\begingroup$ It makes me think. I think was confused between the force applied on current and force applied on current carrying conductor. So in a current carrying conductor, force should be applied on electrons flowing..? $\endgroup$ Dec 25, 2023 at 5:02
  • $\begingroup$ The magnetic Lorentz force acts on the moving charges in the conductor. It 'tries' to push them out of the side of the conductor. The charges are restrained from leaving the conductor by essentially electrostatic force from the positive ions, acting 'inwards'. The Newton's third law partner to this force is an outward force on the ions. This is the force that the wire experiences. It is equal to the sum of the Lorentz forces on the free electrons. [It's a bit like me standing on a weighing machine: the downward force applied to the machine is not the pull of gravity on me but the N3L partner $\endgroup$ Dec 25, 2023 at 9:17
  • $\begingroup$ to the weighing machine's upward contact force on me, which is equal and opposite to the pull of gravity on me!] $\endgroup$ Dec 25, 2023 at 9:20
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The correct formula to find the force due to the ${\bf B}$ field on a single charge in the beam is: $${\bf F} = q{\bf v} \times {\bf B}$$ where $q$ is the charge of the particle. Let us, say that the right is the positive $x$-direction and up is the positive $y$-direction. This implies positive $z$-direction is out of the to preserve right handedness (i.e. preserve ${\bf \hat{x}} \times{\bf \hat{y}}={\bf \hat{z}}$) as is common in this context. With this choice, ${\bf v} = v{\bf \hat{x}}$ where $v>0$ and ${\bf F} = -F{\bf \hat{y}}$ where $F>0$. We know this because the electron is being deflected down. The $q = -|e|$ since the charged particle is an electron. Plugging this in we have, $$-F{\bf \hat{y}} = -v|e|{\bf \hat{x}}\times{\bf B}$$ or $${\bf \hat{y}} = k{\bf \hat{x}}\times{\bf B}$$ for some constant $k>0$. Using the right hand rule, we can see that the only choice of direction for ${\bf B}$ is into the page, i.e. ${\bf B} = -B{\bf \hat{z}}$ for $B>0$. This follows from the fact that $${\bf \hat{x}} \times {\bf \hat{z}} = {\bf -\hat{y}}$$ with the coordinate system chosen.

Conceptual answer, no maths

The starting point is to consider that a negative charge travelling to the right (as in the picture) will be feel a force by a magnetic field

  1. Up, if the magnetic field is out of the page or,
  2. Down, if the magnetic field is into the page.

This comes from the Lorentz force law but can be takes as an axiom here.

Since the trajectory of the electron shows the electron being deflected down, we know the force due to the magnetic field has to be down. By the above, the field is into the page.

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    $\begingroup$ er, I'm in highschool (not the us one) and so I can't understand all these equations. Can you please explain them in easier words? $\endgroup$ Dec 24, 2023 at 17:40
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    $\begingroup$ I will leave the answer as it is more useful to more people. I will make an addendum to your specific knowledge level. In the future, please specify this in the question as it is a site rule. Is the first equation (in some form) familiar? $\endgroup$
    – JohnA.
    Dec 24, 2023 at 17:42
  • $\begingroup$ Yes, I have seen first equation many time but still haven't used it (or read about it properly.) $\endgroup$ Dec 25, 2023 at 4:53

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