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According to the comments below the answer of the question about Experiments with Alpha particles in a homogeneous magnetic field an inhomogeneous magnetic field will deflect neutrons. Why a homogeneous magnetic field will not deflect neutrons?

The use of an inhomogeneous magnetic field in the Stern-Gerlach-experiment is to separate particles in two regions and not to get a smear. But IF an inhomogeneous magnetic field will deflect neutrons THAN a homogeneous field has to deflect too?

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HolgerFiedler
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    I'm voting to close this question as off-topic because of insufficient research effort. – CuriousOne Jul 20 '16 at 20:35
  • It is not clear at all why you think that things at are deflected in inhomogeneous fields should also be deflected by homogeneous fields. It is not difficult to find out that the force on a magnetic moment depends on the spatial derivative of the magnetic field, which vanishes in the homogeneous case. – ACuriousMind Jul 21 '16 at 13:16

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Lets treat this classically, for simplicity. Neutron has no charge. However, it has a intrinsic magnetic dipole moment $\mathbf\mu$. The force of the magnetic field applied to this dipole: $$ \mathbf F = \nabla(\mathbf\mu\cdot\mathbf B) $$

Since you are assuming the magnetic field does not depend on space (its homogenous), then its "gradient" will be zero. That is: $$ F_i = \sum_k\frac{\partial}{\partial x_i}\left(\mu_k B_k\right) = \sum_k\mu_k\frac{\partial B_k}{\partial x_i} = 0,\quad \mbox{since}\quad \frac{\partial B_k}{\partial x_i} = 0 $$

Then, force is zero. Thus, no deflection.

Physicist137
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  • The gradient is made in the direction perpendicular to the trajectory of the particle? – HolgerFiedler Jul 20 '16 at 20:06
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    @HolgerFiedler No. This is not $\mathbf F = q\mathbf v\times\mathbf B$. There is no velocity there, so, force is independent on trajectory (velocity). Force only depends on its position in space (ie, of the magnetic field in that position). The gradient is taken of the dot product of magnetic moment and external magnetic field, as its shown in answer. – Physicist137 Jul 20 '16 at 20:10
  • So independent from a homogeneous or an inhomogeneous field in the direction of the particles trajectory the gradient is zero. – HolgerFiedler Jul 20 '16 at 20:19
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    @HolgerFiedler What? Why? Well, if field is inhomogeneous, then B is not constant spatially. Thus $\frac{\partial B_k}{\partial x_i}\ne 0$. Thus, force will not be zero. – Physicist137 Jul 20 '16 at 20:32
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    I was going to write an answer almost identical to yours. It is necessary to have a nonconstant magnetic field. – Lawrence B. Crowell Jul 20 '16 at 21:49
  • @Physicist137 According to this link http://physicsinsights.org/force_on_dipole_1.html (thanks to tob) "my" is the dipole moment of a current. This is not applicable to a neutron. – HolgerFiedler Jul 21 '16 at 06:54
  • @HolgerFiedler Nop. Its applicable to neutron. On classical physics, the source of the B-Field is the current. No current, no B-Field. The multipole expansion from a current distribution leads to magnetic dipoles, quadrupoles, etc. So, magnets, Earth, and neutron, only generates magnetic field, because there is current somewhere inside them generating it. There is no other way. (This is classical). However, what you could have said, is that this is not applicable to a neutron, because neutron is a quantum particle. However, the quantum treatment will lead to the same result: No deflection. – Physicist137 Jul 21 '16 at 19:18
  • Following your statement that magnetic dipole moment comes only from a current how this is applicable to an electron? More than this, the smallest unit of a n electric current is the electron, so how the electrons magnetic dipole moment will be induced by a current inside the electron? – HolgerFiedler Jul 21 '16 at 21:13
  • @HolgerFiedler Simple. Moving electron, means "current", which means, magnetic field. About second question, simple too: The electron is a quantum particle. Not a classical one, thus my classical explanation cannot apply. My whole answer was classical..... – Physicist137 Jul 21 '16 at 22:01