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Force on magnetic field and velocity

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anillatoo
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When the force on a charge moving in a circular motion in a magnetic field is doubled, what is the effect on its velocity? I thought since Force = Bqv, the force is doubled but turns out that I am wrong because the Force is also mv^2/r and it is quadrupled. So, how do I know which equation to use?
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JohnnyDavidson
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(Original post by anillatoo)
When the force on a charge moving in a circular motion in a magnetic field is doubled, what is the effect on its velocity? I thought since Force = Bqv, the force is doubled but turns out that I am wrong because the Force is also mv^2/r and it is quadrupled. So, how do I know which equation to use?
Hi, is this an exert from a question, and does the mark scheme give an answer of velocity becomes 4x? I make it that velocity is doubled, when force is doubled, for both equations

I think you're getting getting confused between velocity & force. By using the centripetal force equation;

If the force is doubled, then the velocity will be doubled (provided radius and mass remain constant), see photo attachment Name: image.jpg Views: 161 Size: 248.6 KB

If the velocity is doubled, then the force will be quadrupled (provided radius and mass remain constant), since the velocity is squared (2^2 = 4)
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anillatoo
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(Original post by JohnnyDavidson)
Hi, is this an exert from a question, and does the mark scheme give an answer of velocity becomes 4x? I make it that velocity is doubled, when force is doubled, for both equations

I think you're getting getting confused between velocity & force. By using the centripetal force equation;

If the force is doubled, then the velocity will be doubled (provided radius and mass remain constant), see photo attachment Name: image.jpg Views: 161 Size: 248.6 KB

If the velocity is doubled, then the force will be quadrupled (provided radius and mass remain constant), since the velocity is squared (2^2 = 4)
Sorry, yea, I got confused. So, when do I use F = Bqv?
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Konanabanana
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(Original post by JohnnyDavidson)
Hi, is this an exert from a question, and does the mark scheme give an answer of velocity becomes 4x? I make it that velocity is doubled, when force is doubled, for both equations

I think you're getting getting confused between velocity & force. By using the centripetal force equation;

If the force is doubled, then the velocity will be doubled (provided radius and mass remain constant), see photo attachment Name: image.jpg Views: 161 Size: 248.6 KB

If the velocity is doubled, then the force will be quadrupled (provided radius and mass remain constant), since the velocity is squared (2^2 = 4)
Centripetal force does not have an equation..... mv^2/r is not centripetal force it is the resultant of the forces, teachers who teach it as centripetal will cost students marks in the exams.
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JohnnyDavidson
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(Original post by anillatoo)
Sorry, yea, I got confused. So, when do I use F = Bqv?
You could have used either here, both will give velocity doubled.F = bqv
if force doubled (times both side by 2): 2F = 2bqv
Therefore, provided b and q remain constant, velocity will be doubled.

(Original post by Konanabanana)
Centripetal force does not have an equation..... mv^2/r is not centripetal force it is the resultant of the forces, teachers who teach it as centripetal will cost students marks in the exams.
Yes that is true, perhaps I didn't quite explain it fully, though I am only an a2 student myself so not sure exactly on the definition lol.

Centripetal force is the resultant force, which can be applied whenever an object is in uniform circular motion. In this case the only force acting on the object/charge is the magnetic force, so this is therefore the resultant force, which is the centripetal force.

So yes centripetal force is never a separate force, it is always "provided" by something else (in this case only the magnetic force, if there were more forces, it would be the resultant force of those).

So if the object is in uniform circular motion, the velocity (magnitude of), radius of orbit, and mass, can be found using the equation: F = (mv^2) / r, taking F as the resultant force.
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Konanabanana
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(Original post by JohnnyDavidson)
You could have used either here, both will give velocity doubled.F = bqv
if force doubled (times both side by 2): 2F = 2bqv
Therefore, provided b and q remain constant, velocity will be doubled.



Yes that is true, perhaps I didn't quite explain it fully, though I am only an a2 student myself so not sure exactly on the definition lol.

Centripetal force is the resultant force, which can be applied whenever an object is in uniform circular motion. In this case the only force acting on the object/charge is the magnetic force, so this is therefore the resultant force, which is the centripetal force.

So yes centripetal force is never a separate force, it is always "provided" by something else (in this case only the magnetic force, if there were more forces, it would be the resultant force of those).

So if the object is in uniform circular motion, the velocity (magnitude of), radius of orbit, and mass, can be found using the equation: F = (mv^2) / r, taking F as the resultant force.
Forces always will have components, meaning there is always more than one force acting on them. Also centripetal force is not a force.
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JohnnyDavidson
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(Original post by Konanabanana)
Forces always will have components, meaning there is always more than one force acting on them. Also centripetal force is not a force.
How would you explain it then?
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anillatoo
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How is it not a Force?! Anyway, I am still confused about the effect of Force on velocity in a magnetic field; on one hand it is proportional (F=Bqv) and on another hand, it is proportional to the square(F= mv^2/r)
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Student403
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(Original post by Konanabanana)
Forces always will have components, meaning there is always more than one force acting on them. Also centripetal force is not a force.
Not necessarily in our context of A2...
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Absent Agent
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(Original post by anillatoo)
When the force on a charge moving in a circular motion in a magnetic field is doubled, what is the effect on its velocity? I thought since Force = Bqv, the force is doubled but turns out that I am wrong because the Force is also mv^2/r and it is quadrupled. So, how do I know which equation to use?
You cannot really answer the question until a few things are assumed. However, since the question demands the effect of velocity when the magnetic force is doubled, we need to assume that magnetic field strength, charge and radius of orbit are remained constant. Therefore, you cannot use the magnetic force equation, F=qvB, because the force in this equation is the dependent variable, meaning the force on a charge particle varies as a result of a change in one or more of the variables q, v, B, but not vice versa.

On the other hand, the equation F= \dfrac{mv^2}{r} *could* express the effect of velocity of the charged particle in the magnetic field if the magnetic force was to be doubled (assuming the mass and radius of the orbit remain constant). Hence,

\dfrac{m}{r}=\dfrac{F}{v^2}=C, where C is a constant. Then we have,

\dfrac{F_1}{v^2_1}=\dfrac{F_2}{v ^2_2}. Rearranging for v_2 we get,

v_2=\sqrt{\dfrac{F_2v^2_1}{F_1}} =\sqrt{\dfrac{(2F_1)v^2_1}{F_1}} =\sqrt{2}v_1.
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anillatoo
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(Original post by Absent Agent)
You cannot really answer the question until a few things are assumed. However, since the question demands the effect of velocity when the magnetic force is doubled, we need to assume that magnetic field strength, charge and radius of orbit are remained constant. Therefore, you cannot use the magnetic force equation, F=qvB, because the force in this equation is the dependent variable, meaning the force on a charge particle varies as a result of a change in one or more of the variables q, v, B, but not vice versa.

On the other hand, the equation F= \dfrac{mv^2}{r} *could* express the effect of velocity of the charged particle in the magnetic field if the magnetic force was to be doubled (assuming the mass and radius of the orbit remain constant). Hence,

\dfrac{m}{r}=\dfrac{F}{v^2}=C, where C is a constant. Then we have,

\dfrac{F_1}{v^2_1}=\dfrac{F_2}{v ^2_2}. Rearranging for v_2 we get,

v_2=\sqrt{\dfrac{F_2v^2_1}{F_1}} =\sqrt{\dfrac{(2F_1)v^2_1}{F_1}} =\sqrt{2}v_1.
Thanks, that was very helpful.
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lerjj
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There seems to be some confusion between a few things:

Firstly, increasing the magnetic field will increase the force on the electron. It will not however, change it's speed as magnetic forces do no work. What it will do is cause it to move in a tighter circle.

Second - the term centripetal force refers to any force causing circular motion. It is not an 'extra' force you get when moving in a circle. It just happens that in order to turn, you need to apply a force.

Thirdly - you can't use equations backwards! If you double the velocity of the particle, then the force will double (again, will go in a smaller circle) but the reverse is not true because you can't just arbitrarily "double the force" - what you can do is double the field. Same with v^2/r - if you increase v you don't have to increase the force as you might simply stop going in a circle (not in this case though).
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0le
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(Original post by Konanabanana)
Forces always will have components, meaning there is always more than one force acting on them. Also centripetal force is not a force.
Why is the centripetal force not a force? Its a name for any force that keeps an object in circular motion, whether its cause is tension or friction (or something else?). The centrifugal force is the one that is fictitious and comes about depending on reference frame.
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