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##.
 

1. What is a solenoid?

A solenoid is a type of electromagnet consisting of a coil of wire that is tightly wound in a helix shape. It is used to generate a magnetic field when an electric current is passed through it.

2. How is an electric field generated by a solenoid?

When an electric current flows through a solenoid, it creates a magnetic field around the coil. This magnetic field, in turn, generates an electric field in the space surrounding the solenoid.

3. What factors affect the strength of the electric field generated by a solenoid?

The strength of the electric field generated by a solenoid is affected by the number of turns in the coil, the current flowing through the coil, and the material of the core inside the coil. The closer the turns are to each other, the stronger the field will be.

4. What is the direction of the electric field generated by a solenoid?

The direction of the electric field generated by a solenoid is parallel to the axis of the coil. This means that the electric field lines will be running in a straight line from one end of the coil to the other.

5. How is the electric field strength of a solenoid calculated?

The electric field strength of a solenoid can be calculated using the equation E = N x I / L, where N is the number of turns in the coil, I is the current flowing through the coil, and L is the length of the solenoid. This equation is known as Ampere's law.

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