Solenoid flux-emf-current problem

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In summary, the problem involves a solenoid inductor with an emf of 0.20 V and a steady current of 0.10 A producing a flux of 5.0 microWb per turn. The goal is to determine the number of loops in the solenoid, and this can be done by combining the equations \phi=BA and B=\frac{\mu NI}{l}. By dividing both sides of the equation \phi=\frac{\mu NIA}{l} by N, the ratio of \frac{A}{l} can be found. Finally, the equation E=\frac{d}{dt}{\phi} or E=\frac{\phi}{t} can be used to solve for
  • #1
crbnmnxd3
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Homework Statement


A solenoid inductor has an emf of 0.20 V when the current through it changes at the rate 10.0 A/s. A steady current of 0.10 A produces a flux of 5.0 microWb per turn.

The number of loops in the solenoid is the value which needs to be found.

Homework Equations





The Attempt at a Solution


I have been going through this problem for a good hour, and am simply stuck. Clearly there are two different cases which are presented for the same solenoid, which clearly means that each case needs to be plugged into an equation and these two equations must be related through an unknown variable. Please state any equations you may use in the solving of this problem, I have two pages of fruitless work in front of me.
 
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  • #2
[tex]\phi=BA[/tex]

[tex]B= \frac{\mu NI}{l}[/tex]

N=No. of turns
l=length.

You'll need those 2.
 
  • #3
The problem I keep running into with the first equation you listed is that I do not know what the radius of the solenoid is, and as a result I am unable to determine the surface area of the solenoid.
 
  • #4
Combining the 2 equations you'll get

[tex]\phi = \frac{\mu NIA}{l}[/tex]


Dividing both sides by N


[tex]\frac{\phi}{N}=\frac{\mu IA}{l}[/tex]

Can you use the second line of the question to find a value for the ratio of [itex]\frac{A}{l}[/itex]?


Then use the fact that

[tex]E=\frac{d}{dt}{\phi}[/tex]

or simply

[tex]E=\frac{\phi}{t}[/tex]
 

1. What is a solenoid in relation to flux, EMF, and current?

A solenoid is a coil of wire that produces a magnetic field when an electric current is passed through it. This magnetic field can induce an electromotive force (EMF) and a current in the wire.

2. How does the flux in a solenoid affect the EMF and current?

The flux in a solenoid is directly proportional to the induced EMF and current. As the flux increases, so does the EMF and current, and vice versa.

3. What is the formula for calculating the EMF induced in a solenoid?

The formula for calculating the EMF induced in a solenoid is EMF = -N(dΦ/dt), where N is the number of turns in the solenoid and dΦ/dt is the rate of change of magnetic flux.

4. How is the direction of the induced current in a solenoid determined?

The direction of the induced current in a solenoid is determined by Lenz's law, which states that the induced current will flow in a direction that opposes the change in magnetic flux that caused it.

5. What factors affect the EMF and current in a solenoid?

The EMF and current in a solenoid are affected by the number of turns in the solenoid, the rate of change of magnetic flux, and the resistance of the wire in the solenoid. These factors can be manipulated to control the strength and direction of the induced current.

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