1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: How to find magnetic flux density at center and ends of solenoid

  1. Aug 4, 2012 #1
    1. The problem statement, all variables and given/known data

    A solenoid has a radius of 2mm and a length of 1.2cm. If the # of turns per unit length is 200 and the current is 12A, calculate the magnetic flux density at a) the center and b) the ends of the solenoid

    2. Relevant equations

    The biot-savart law:

    [itex]\vec{B} = \frac{\mu_0}{4 \pi} \ \int \frac{ I \ \vec{dl}\times \hat{r}}{r^2} \ \text{ or } \ \frac{\mu_0}{4 \pi} \ \int \frac{ I \ \vec{dl}\times \vec{r}}{r^3}[/itex]

    More importantly, my textbook has taken me through a few examples to wind up with the following general results for a circular loop of current with radius b.

    In general, on the axis of a current-carrying loop:

    [itex]\vec{B} = \frac{\mu_0 \ I \ {b^2}}{2 \ ({{b^2} + {z^2}})^{3/2}} \hat{z}[/itex]

    By setting z=0, we can obtain the magnetic flux density at the center of the loop as

    [itex]\vec{B} = \frac{\mu_0 \ I}{2 \ b} \hat{z}[/itex]

    I believe i am expected to use only these equations to solve the problem.

    3. The attempt at a solution

    We have 200 turns per unit length, thus we have a total of N = (200)(1.cm) = 2.4 turns

    I'm guessing what I have to do is sum the B from each loop / turn. I know that for the loop that is in the plane of the center, the magnetic flux density is given by

    [itex]\vec{B} = \frac{\mu_0 \ I}{2 \ b} \hat{z}[/itex]

    and that for all the others, the magnetic flux density is given by

    [itex]\vec{B} = \frac{\mu_0 \ I \ {b^2}}{2 \ ({{b^2} + {z^2}})^{3/2}} \hat{z}[/itex]

    where b is the radius and is given.

    I don't know what to put in for z in the above equation. Furthermore i'm not sure what to do for part b).

    I think there must be a simplifying assumption somewhere. Either that or potentially somehow use the general equation for the Biot-Savart law and integrate along the path of the helix
    Last edited: Aug 4, 2012
  2. jcsd
  3. Aug 4, 2012 #2


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    Break the solenoid into thin rings of width dz and use [itex]B= \frac{\mu_0 \ I \ {b^2}}{2 \ ({{b^2} + {z^2}})^{3/2}} \hat{z}[/itex] for the ring. You'll need to think about how to express I for the ring. Then consider how you would sum up the contributions to B from all the rings.
  4. Aug 5, 2012 #3
    Thanks for the response. So I've integrated that expression over the length of the solenoid (from z=0 to z=1.2cm in this case) and i've expressed I as

    [itex] I = 12 \ N [/itex]

    where N is the number of turns per unit length, in this case 200.

    Note that I haven't covered Gauss' law for magnetic fields yet, and the above expression for I seems to be derived from that law everywhere I look. So I don't really understand why that expression above is correct(yet), but it seems to be so.


    [itex] I = 12 \ N = 2400 [/itex]

    [itex] \vec B = \frac{\mu_0 \ I \ {b^2}}{2} \int_{0}^{1.2cm} \frac{dz}{({b^2}+{z^2})^{(3/2)}} [/itex]

    [itex] \equiv \frac{2400 \ \mu_0}{2} [\frac{z}{\sqrt{{b^2}+{z^2}}}] [/itex]

    and the above expression is evaluated from z=1.2cm to z=0 (wasn't sure how to fit that in latex above)

    Is this correct? Also, i'm assuming that the magnetic field will be the same everywhere inside the solenoid and so the answer for part a) and b) is the same.

    Any help would be greatly appreciated, i'd really like to understand this particular question
  5. Aug 5, 2012 #4


    User Avatar
    Homework Helper
    Gold Member
    2017 Award

    A few things need to be corrected here. Let's take the case of finding B at the center of the solenoid. In this case, it's nice to take the origin of the z axis at the center of the solenoid. The solenoid will then extend from z = -L/2 to z = L/2, where L is the length of the solenoid.

    Consider a thin ring of the solenoid of width dz and center located at z. You can use your expression for the magnetic field of a ring to write an expression for the magnetic field produced at z = 0 by this ring. When you do this, you'll need to consider how much current is in the ring. This will be the current in each winding (I) multiplied by the number of windings in a length dz.

    Considering all of this, try to set up the correct expression for the integral that will give you the total field at the origin.

    The case of finding the field at one end is very similar except I would suggest that you then pick the origin of the z-axis at one end.

    The magnetic field is not uniform along the z-axis of a solenoid of finite length, so you cannot assume the answer for B at one end will be the same as at the center.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook