Electric field generated by solenoid

In summary: The electric field at any point is determined by the vectors ##\vec E, \vec B## and the direction of the current.
  • #1
quarkyphysicsgirl
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Homework Statement
"A long solenoid has a radius of 2.08 cm and 1070 turns per meter. Over a certain time interval the current varies with time according to the expression I = 2.80t, where I is in amperes and t is in seconds. Calculate the electric field 4.54 cm from the axis of the solenoid."
Relevant Equations
E=(1/2r)(alpha)R^2(muo)Ioe^-(alpha)t.
The formula we are given is E=(1/2r)(alpha)R^2(muo)Ioe^-(alpha)t.

However, I am struggling to figure out what each of the symbols stands for in the formula...can someone help me out? Like super confused on what alpha is in this case.
 
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  • #2
quarkyphysicsgirl said:
Homework Statement:: "A long solenoid has a radius of 2.08 cm and 1070 turns per meter. Over a certain time interval the current varies with time according to the expression I = 2.80t, where I is in amperes and t is in seconds. Calculate the electric field 4.54 cm from the axis of the solenoid."
Relevant Equations:: E=(1/2r)(alpha)R^2(muo)Ioe^-(alpha)t.

The formula we are given is E=(1/2r)(alpha)R^2(muo)Ioe^-(alpha)t.

However, I am struggling to figure out what each of the symbols stands for in the formula...can someone help me out? Like super confused on what alpha is in this case.
Well well, hello @quarkyphysicsgirl,
:welcome: ##\qquad## !​

You are struggling, but I am too ! The variables in the problem statement don't even appear in the formula you were given !
You sure there isn't a huge mixup going on here ?

Not only that, but the problem statement is imposssible: 4.54 cm from the axis ? Where ? In New York or in Cincinnati ?

Back to the drawing board (or to teacher). It may also help to consult your notes and/or textbook.

Finally, PF has a bunch of rules/guidelines -- well worth reading :wink:

and - as a PS - check out ##\LaTeX##

##\ ##
 
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  • #3
quarkyphysicsgirl said:
The formula we are given is E=(1/2r)(alpha)R^2(muo)Ioe^-(alpha)t.

However, I am struggling to figure out what each of the symbols stands for in the formula...can someone help me out? Like super confused on what alpha is in this case.
That formula doesn't apply to this problem.

In this problem, you have a changing current and therefore a changing magnetic field produced by the solenoid. A changing magnetic field induces an electric field. The problem can be solved by a straightforward application of Faraday's law.

BvU said:
Not only that, but the problem statement is imposssible: 4.54 cm from the axis ? Where ? In New York or in Cincinnati ?
4.54 cm from the axis of the solenoid. I'm not sure why you think there's a problem here.
 
  • #4
@quarkyphysicsgirl What is the context of this problem? Is it from an introductory physics course? When they say "long solenoid" it is supposed to be an indication that the formula for ideal solenoid is used. The point indicated (4.54 cm) it is outside the solenoid (radius is 2.08 cm) and the magnetic field is ideally zero at this point. As it is at all points outside the solenoid. Something seems to be wrong. Or weird.
 
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Likes berkeman and BvU
  • #5
Just because ##\vec B=0## outside the solenoid doesn't mean ##\vec E=0##.
 

FAQ: Electric field generated by solenoid

What is the electric field generated by a solenoid?

The electric field generated by a solenoid is typically induced when there is a change in the magnetic flux through the solenoid. According to Faraday's Law of Induction, a changing magnetic field inside the solenoid induces an electric field that forms closed loops around the solenoid's axis.

How is the electric field inside a solenoid calculated?

The electric field inside a solenoid can be calculated using Faraday's Law of Induction. If the magnetic flux through the solenoid changes, the induced electric field (E) can be determined by the rate of change of the magnetic flux (Φ) through the solenoid: E = -dΦ/dt. The electric field forms concentric circles around the axis of the solenoid.

Does a steady current in a solenoid generate an electric field?

No, a steady current in a solenoid does not generate an electric field. A steady current produces a constant magnetic field, and according to Faraday's Law, only a changing magnetic field can induce an electric field. Therefore, a constant current results in a constant magnetic field and no induced electric field.

What role does the number of turns in a solenoid play in the electric field generated?

The number of turns in a solenoid affects the magnetic flux linkage, which in turn influences the induced electric field. A solenoid with more turns will have a greater magnetic flux for a given current, and thus, a greater rate of change in magnetic flux can induce a stronger electric field when the current changes.

Can the electric field generated by a solenoid be used for practical applications?

Yes, the electric field generated by a solenoid can be used in practical applications such as inductors, transformers, and electromagnetic induction devices. These applications rely on the principle of changing magnetic fields inducing electric fields to transfer energy or signal from one circuit to another.

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