## \qquad## !That is possible, yes. Example: charged particle in a magnetic field. Google Lorentz force.ahmadphy said:
The force itself does not depend on ##{\rm d} t##.ahmadphy said:
Thank you...I know lorentz force...what I want from the question is to understand d'alembert's principle and what I meant by inertial force of the particle due to its kinetic energy is that when a force acts on a moving body it will follow some path I can have the same path if I assumed the particle is at rest and transformed the kinetic energy to force then apply the same force as the situation when it was movingBvU said:Hello @ahmadphy ,
That is possible, yes. Example: charged particle in a magnetic field. Google Lorentz force.The force itself does not depend on ##{\rm d} t##.
You can say ##{\rm d }t={\rm d} x/v## because that's the definition of ##v##.
[ edit ] mathematicians may frown on this...I don't know what 'the inertial force of the particle' means.
##\ ##
Also when I asked if I can say that dt=dx/v is it always mathematically correct?ahmadphy said:
Well, ##v=0## requires an exceptionahmadphy said:
I really still don't understand... Do you have a reference or an example?ahmadphy said:
The electromagnetic force experienced by a charged particle depends on its velocity due to the Lorentz force law. This law states that the total force on a particle is the sum of the electric force and the magnetic force, where the magnetic force is velocity-dependent. Specifically, the magnetic force is given by \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \), where \( q \) is the charge, \( \mathbf{v} \) is the velocity, and \( \mathbf{B} \) is the magnetic field.
A stationary charged particle is unaffected by a magnetic field because the magnetic component of the Lorentz force depends on the velocity of the particle. If the particle's velocity \( \mathbf{v} \) is zero, the magnetic force \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \) will also be zero. Thus, only the electric field can exert a force on a stationary charged particle.
The cross-product in the magnetic force equation \( \mathbf{F} = q(\mathbf{v} \times \mathbf{B}) \) indicates that the force is perpendicular to both the velocity of the particle and the magnetic field. This perpendicular nature of the force causes the particle to move in a circular or helical path when only a magnetic field is present, rather than in a straight line.
No, a magnetic field cannot do work on a charged particle. The force exerted by a magnetic field is always perpendicular to the velocity of the particle, meaning that it changes the direction of the particle's motion but not its speed. Since work is defined as the force component in the direction of displacement, and there is no component of the magnetic force in the direction of the particle's velocity, the magnetic field does no work.
In a combined electric and magnetic field, the trajectory of a charged particle is influenced by both fields. The electric field exerts a force in the direction of the field, affecting the particle's speed and direction. The magnetic field exerts a force perpendicular to the particle's velocity, altering its direction without changing its speed. The resulting trajectory can be complex, often resulting in a helical or spiraling path depending on the initial velocity and the orientations of the fields.