How does one calculate the flux density B at end of solenoid?

[Nile Anderson]

Homework Statement

I came across a recent problem that asked me to calculate the flux density at the end of a solenoid. I was given the current 3 A , the number of turns per unit length , 12 cm^-1 and the using the permeability of free space as 4π × 10^-7

Homework Equations

The equation I used for this is B=μnI , where n is the number of turns per unit length ( in per meter) and I current.

The Attempt at a Solution

I got 144π × 10^-5, this however was not an option and so I was wondering if there were some other factor I had to account for as it is the end of the coil and not for the midpoint for which the equation is defined.

 

[NascentOxygen]

You need to show your working so helpers can first discount that you may have made a simple mistake in the mathematics.

 

[Nile Anderson]

You need to show your working so helpers can first discount that you may have made a simple mistake in the mathematics.

With all due respect sir, I have no problems using a calculate but ok , never the less:
You need to convert the 12 per cm to per m which gives you 12/10^-2=1200
B= 4π×10^-7 ×1200 × 3 = 144π×10^-5

 

[gneill]

I got 144π × 10^-5, this however was not an option and so I was wondering if there were some other factor I had to account for as it is the end of the coil and not for the midpoint for which the equation is defined.

Presumably this is a multiple choice style of question?

Can you quote the problem exactly along with the selection of answers?

It sometimes happens that a problem is "refreshed" with a slight change in values in order to make it "new" (so last year's answers can't be copied). Occasionally the "refresh" is imperfect: either the new answer is not inserted in the list of answers or the person making the change made an error in calculation...

 

[NascentOxygen]

B=µnI gives the flux density midway between the two ends of the solenoid.

Follow the link in this thread to obtain the formula for B near the end of a solenoid. If I read it correctly, you'll find that right at the end they'll differ by a factor of approx. 2

 

[Nile Anderson]

The question said "What is the magnetic flux density near the end of a solenoid that has 12 turns per centimeter and carrying a current of 3A? (Assume μ_o =4π×10^-7)
Answers
A) 144π×10^-7) T
B) 1.44π×10^-7) T
C) 144π μT
D) 144π T

Presumably this is a multiple choice style of question?

Can you quote the problem exactly along with the selection of answers?

It sometimes happens that a problem is "refreshed" with a slight change in values in order to make it "new" (so last year's answers can't be copied). Occasionally the "refresh" is imperfect: either the new answer is not inserted in the list of answers or the person making the change made an error in calculation...

 

[NascentOxygen]

Answers
A) 144π×10^-7) T
B) 1.44π×10^-7) T
C) 144π μT
D) 144π T

Looks like correct answer is: (E) none of the above

 

[Nile Anderson]

Looks like correct answer is: (E) none of the above

Lol yes , a reasonable conclusion

 

[theodoros.mihos]

The two factors are limits for very long and very short solenoid relative to ring radius, so I think that you ask for the assumptions to use on your class level.

 

[rude man]

The question said "What is the magnetic flux density near the end of a solenoid that has 12 turns per centimeter and carrying a current of 3A? (Assume μ_o =4π×10^-7)
Answers
A) 144π×10^-7) T
B) 1.44π×10^-7) T
C) 144π μT
D) 144π T

Length of solenoid? 1cm? 1m?

 

[rude man]

If length >> radius, answer would be close to A/2.
If length << radius, B = μ₀IR/2
which is of no help since radius R is not given.
What you do is sum the B fields due to each wire turn using Biot-Savart. This is most readily done by the appropriate integral, so what you're doing here is assuming a differentially small wire spacing then integrating over distance from the solenoid's end (x=0) to x=infinity, and pre-multiplying by 1/(actual wire spacing).

 

[Dadface]

Imagine two exacly identical coils carrying the same current in the same direction. B along the axis of each coil is given by by Nile in post one. Let B at the end points of the coils have the value Be. Now imagine that the two coils are pushed together end to end so as to make a longer coil. B along the axis will have the same value as before. Now consider the midpoint of the axis of this coil, where two ends of the smaller coils meet. B at this point can be considered to be the sum of B at the two ends of the smaller coils and therefore B = 2Be. In other words B at the end points of a long coil is half the value of B within the coil.

 

[theodoros.mihos]

See this