Electromagnetic force on a positively charged particle

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The discussion focuses on deriving the net force acting on a positively charged particle due to electromagnetic fields. Participants emphasize the importance of expressing the vectors involved in unit vector notation. The Lorentz force equation is highlighted as a key tool for calculating the net force. The magnitude of the force can be determined by taking the square root of the sum of the squared components. This approach is essential for accurately analyzing the effects of the electromagnetic force on the charged particle.
Sat-P
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Homework Statement
It's not a question, but this is something I tried to derive using the Lorentz force equation on a charged particle.
Relevant Equations
F= qE + qv×B
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Please state what exactly you are trying to derive.
 
Magnitude of the net force acting on the charged particle due to both fields
 
Sat-P said:
Magnitude of the net force acting on the charged particle due to both fields
Then I suggest that you write each of the vectors involved in unit vector notation, add them to get the Lorentz force as per the equation you posted and finally find its magnitude. It will be the square root of the sum of the components squared and will depend on how you chose the three vectors.
 
Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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