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The setup is just a circular loop with a capacitor C on the right side (i.e. at 3 o'clock) and a resistor R on the left side (at 9 o'clock), completely inside a uniform magnetic field. There is no battery, etc; just the emf induced by the magnetic field.
A uniform magnetic field (directed INTO the page) decreases at a constant rate dB/dt = K, where K is a positive constant. A circular loop of wire of radius a containing a resistance R and a capacitance C is placed with its plane normal to the field. (a) Find the charge Q on the capacitor when it is fully charged. (b) Which plate is at the higher potential. (c) Discuss the force that causes the separation of charges.
I have a pretty low level of confidence about this, but here goes...
for part (a), given that
[tex]\frac{dB}{dt} = K [/tex]
The magnetic flux at any point in time is
[tex]\Phi_{B} = \pi r^2B [/tex]
so the induced emf is
[tex]\epsilon = \frac {d\Phi_B}{dt} = \pi r^2\frac{dB}{dt} = \pi r^2K
[/tex]
so wouldn't the charge on the capacitor when "fully charged" be given by
[tex]Q = C\epsilon = \pi r^2KC
[/tex]
??
But then what role does the resistance play? (I think that the resistance affects the time it takes to charge the capacitor, but not the ultimate amount of charge. So, was the resistor placed in this question just to confuse?)
for (b):
Since the magnetic field is directed into the page and is decreasing, the induced current should be such that its magnetic field is in the same direction as the original field. Thus I'm thinking that the flow of the charging current should be clockwise, resulting in a higher potential on the upper plate. Is this correct?
for (c):
(The textbook is pretty vague on this ... at least it seems so to me, so my grasp of the concept is commensurately vague.[b(]
The force exerted on the charges in the loop is the result of a difference in potential induced by the change in the magnetic field. Here, since the direction of the field is into the page, and the field is decreasing, the effect on the charges in the loop is the same as if there was an INCREASING magnetic field directed out of the page, or if the magnetic field was steady and the loop was moving parallel to the field, INTO the page.
[edit] OK, I've decided that that last halfsentence is completely wrong. If the field was CONSTANT, and the loop moved in a direction parallel to the field, i.e., directly into or out of the page but always with the plane of the loop oriented parallel to the page, there would be NO induced current, correct? (Because then the magnetic flux through the loop would be constant, despite the motion of the loop.)
The separation of charges progresses until the force exerted by the magnetic field is balanced by the attractive force due to the electrical field that is produced inside the conductor  is this more or less correct?
This is very confusing. Is this completely unrelated to the magnetic force that would be exerted on a charge moving through the magnetic field, since (at least initially) the charges are not moving? BUT, what about when the charges begin to move in the loop as a result of the induced current? Isn't there then an additional force on the charges since they are moving perpendicular to the field? Do we just ignore that?
A uniform magnetic field (directed INTO the page) decreases at a constant rate dB/dt = K, where K is a positive constant. A circular loop of wire of radius a containing a resistance R and a capacitance C is placed with its plane normal to the field. (a) Find the charge Q on the capacitor when it is fully charged. (b) Which plate is at the higher potential. (c) Discuss the force that causes the separation of charges.
I have a pretty low level of confidence about this, but here goes...
for part (a), given that
[tex]\frac{dB}{dt} = K [/tex]
The magnetic flux at any point in time is
[tex]\Phi_{B} = \pi r^2B [/tex]
so the induced emf is
[tex]\epsilon = \frac {d\Phi_B}{dt} = \pi r^2\frac{dB}{dt} = \pi r^2K
[/tex]
so wouldn't the charge on the capacitor when "fully charged" be given by
[tex]Q = C\epsilon = \pi r^2KC
[/tex]
??
But then what role does the resistance play? (I think that the resistance affects the time it takes to charge the capacitor, but not the ultimate amount of charge. So, was the resistor placed in this question just to confuse?)
for (b):
Since the magnetic field is directed into the page and is decreasing, the induced current should be such that its magnetic field is in the same direction as the original field. Thus I'm thinking that the flow of the charging current should be clockwise, resulting in a higher potential on the upper plate. Is this correct?
for (c):
(The textbook is pretty vague on this ... at least it seems so to me, so my grasp of the concept is commensurately vague.[b(]
The force exerted on the charges in the loop is the result of a difference in potential induced by the change in the magnetic field. Here, since the direction of the field is into the page, and the field is decreasing, the effect on the charges in the loop is the same as if there was an INCREASING magnetic field directed out of the page, or if the magnetic field was steady and the loop was moving parallel to the field, INTO the page.
[edit] OK, I've decided that that last halfsentence is completely wrong. If the field was CONSTANT, and the loop moved in a direction parallel to the field, i.e., directly into or out of the page but always with the plane of the loop oriented parallel to the page, there would be NO induced current, correct? (Because then the magnetic flux through the loop would be constant, despite the motion of the loop.)
The separation of charges progresses until the force exerted by the magnetic field is balanced by the attractive force due to the electrical field that is produced inside the conductor  is this more or less correct?
This is very confusing. Is this completely unrelated to the magnetic force that would be exerted on a charge moving through the magnetic field, since (at least initially) the charges are not moving? BUT, what about when the charges begin to move in the loop as a result of the induced current? Isn't there then an additional force on the charges since they are moving perpendicular to the field? Do we just ignore that?
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