Calculating Magnetic Force of Fe+ ion at a Given Point

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To calculate the instantaneous magnetic force on a Fe+ ion traveling at 200 m/sec, the relevant formula is F_m = q(V x B), where V is the velocity vector and B is the magnetic field vector. The angle between the velocity and magnetic field vectors is crucial for determining the force's magnitude and direction. A correction was made regarding the vector cross product, emphasizing that the order matters in calculating the magnetic force. The right-hand rule is commonly used to determine the direction of the magnetic force. Accurate calculations are essential to avoid errors in the direction of the force.
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A Fe+ ion is traveling at 200 m/sec in a direction 20 deg south of west at a point. The magnetic field at that point is 0.4 T directed at 10 deg west of south. What is the instantaneous magnetic force that acts on the Fe is how many Newtons?
 
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The magnetic force on a charge that travels through a magnetic field is:
\vec F_m = q\vec V \times \vec B = q|V||B|\sin \alpha
(\alpha[/tex] being the angle between the field vector and the velocity vector of course.)
 
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Originally posted by Chen
The magnetic force on a charge that travels through a magnetic field is:
\vec F_m = q\vec B \times \vec V = q|B||V|\sin \alpha
(\alpha[/tex] being the angle between the field vector and the velocity vector of course.)
<br /> <br /> I normally would not be picky on such things. However, it should be corrected that the general form of Lorentz force for the magnetic field term is proportional to q(v x B), not q(B x v).<br /> <br /> Since the original question involves the exact vectorial direction of the force, and since cross products are not commutative, this can produce an error in the direction.<br /> <br /> Zz.
 
Of course, I've corrected my post. Thanks for pointing my mistake out. :smile: Usually I just use the right-hand rule to find the direction of the magnetic force.
 
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