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Consider the motion of a charged particle in a uniform magnetic field. $\vec{B} = B_0(-\hat{k})$. Let the initial velocity with which it enters the field be $\vec{v_i} = v_0(-\hat{i})$. It is well known that it follows a circular path of radius $R = \frac{m v_0}{qB_0}$.

  1. Using Work Energy Theorem $$∆K = W_B = \int \vec{F_B}.\vec{v}dt = \int q(\vec{v}×\vec{B}).\vec{v}dt = 0$$ $$ K_f = K_i $$ Therefore speed of the charged particle and radius of the path remains constant.
  2. Using Theory of Electromagnetic Radiation

Direction of velocity changes continuously. Therefore, the charged particle is in accelerated motion. Therefore, it continuously loses energy in the form of electromagnetic radiation. Therefore it must follow a spiral path.

Which of the following is correct?

N.B. I am high school student. So, please limit the answer within high school mathematics.

S Das
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  • Accelerated charges radiate by emitting photons. How exactly does this lead to the electron getting spiralled in to rest is explained here https://physics.stackexchange.com/questions/65339/how-and-why-do-accelerating-charges-radiate-electromagnetic-radiation by @HolgerFiedler – Gravity_CK Jul 25 '21 at 21:19

2 Answers2

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Yes, your second reasoning is correct. Accelerating charges do emit electromagnetic radiation, and what you described is known as synchrotron radiation. This energy loss increases sharply as the particle approaches the speed of light. In fact, this is a limiting factor of the maximum speed that particle accelerators can produce. If no additional energy is provided, then the particle will indeed spiral. The first one is just a simplification, since radiation from accelerating charges are not typically considered at the high-school level.

The power radiated by a point charge is given by the Larmor formula and a derivation of the angular distribution of the radiation (which is slightly more complicated) can be found here.

Vincent Thacker
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  • But, I don't see anything wrong with method 1. Does that mean this is a limitation of work energy theorem? – S Das Jul 25 '21 at 16:26
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    @SDas The electromagnetic energy radiated can be calculated from the Poynting vector. If you integrate this vector over a surface enclosing the setup, you will obtain the energy lost to radiation. This can be added to the work-energy theorem. It's just that the typical work-energy theorem does not include this energy loss. – Vincent Thacker Jul 25 '21 at 16:37
  • @SDas in the derivation for Work Energy theorem they use Newton's second law https://en.wikipedia.org/wiki/Work_(physics)#cite_ref-Young_1-2 and it's a simple fact that Newton's laws don't account for massless particles like photons. – Gravity_CK Jul 25 '21 at 21:58
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There is a formula for the Power of the emitted radiation due to an accelerating charge, the Larmor formula

$$P=\frac{q^2a^2}{6\pi\epsilon_0c^3}$$

where $a$ is the acceleration of the charge also $\frac{Bqv}{m}$.

In calculations the energy lost by radiation can often be ignored, you can put in various numbers to see the size of the effect, but technically you are right, the particle would follow a spiral path.

John Hunter
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