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I read somewhere that a proton takes a semicircular path and exists a magnetic field when it enters. I did some research and that seems to be the case. From my understanding, an electron in a magnetic field goes inside and gets deflected inside it like an inverted C, therefore a proton would have to exit out?

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  • $\begingroup$ No. That would violate conservation of charge. The only difference between them would be the direction of the "turning" and apparent radius depending on the mass of the particle. $\endgroup$
    – Tetradic
    Apr 8, 2015 at 5:46

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The bend radius of a proton (any charged particle) is a function of its charge, the mass, its velocity, and the strength of the magnetic field -

$$r =\frac{mv}{qB}$$

Where velocity is taken perpendicular to the magnetic field.

If "entering a magnetic field" means "going from a region of zero field to a semi infinite region of uniform field" and the entering happens at right angles to the line separating the regions, then the above equation will indeed result in a semicircular path before the proton exits. But that is a very specific set of conditions.

Note that an electron would do the same thing - but since the charge is opposite and the mass much smaller, it has a much tighter bend radius (and while the proton turns one way, the electron turns the other way). For very rapid acceleration (tight turns, high speed) the electron will also lose energy (Bremsstrahlung) so it loses velocity and the turn becomes tighter as it loses speed - so it won't do an exact 180 degree turn before exiting.

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Any charge container has a surrounding magnetic field, and is acted upon by surrounding magnetic fields.

To approach an answer, you'd need to provide much additional detail about the field strength and direction, momentum and direction of the proton/electron, and various other descriptions of the setup.

The best that can be given from what you have here is: small charged particles will be attracted by dissimilar charges and repelled by similar ones. Depending on the particle's initial momentum and direction, its mass, its charge, and the field strength it 'enters', the result could range from minor variations in trajectory, various degrees of deflection, all the way up to reversal ("turning around") or even collisions with other particles leading to other interactions.

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Is vague and undefined, you aren't asking a proper question, and your conclusion of "a proton would have to exit out" cannot be concluded by the previous statements.

Making a number of unreasonable assumptions and trying to guess what you meant, in a homogeneous field of equal strength that reflects electrons of charge -1 and mass 9.10938291 × 10^-31 kilograms and velocity X m/s (momentum=mv), in a given direction, this same field may not reflect a proton of charge +1 and a mass of 1.67262178 × 10^-27 kilograms and velocity X m/s (same velocity), as it carries ~1836 times the momentum.

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