1. Not finding help here? Sign up for a free 30min 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!

A Simple Way to Measure Magnetic Fields

  1. Mar 12, 2008 #1
    1. The problem statement, all variables and given/known data

    A loop of wire is at the edge of a region of space containing a uniform magnetic field B. The plane of the loop is perpendicular to the magnetic field. Now the loop is pulled out of this region in such a way that the area A of the coil inside the magnetic field region is decreasing at the constant rate . That is, [tex]\frac{dA}{dt} = -c[/tex] , with c >0 .

    The induced emf in the loop is measured to be V. What is the magnitude B of the magnetic field that the loop was in?
    Express your answer in terms of some or all of the variables A ,c , and V.

    [​IMG]

    2. Relevant equations

    [tex]\epsilon = | \frac{d\Phi_m}{dt} |[/tex]
    [tex]\Phi_m = AB[/tex]
    [tex]\epsilon = lvB[/tex]

    3. The attempt at a solution

    I get these three hints:

    Hint 1. The formula for the magnetic flux through a loop

    Hint 2. How to take the derivative of the product of two functions

    Hint 3. The formula for the emf induced in a loop (Faraday's law)

    So I know [tex]B = \frac{\Phi_m}{A}[/tex], would I just find the derivative of that?
     
    Last edited: Mar 12, 2008
  2. jcsd
  3. Mar 12, 2008 #2

    rock.freak667

    User Avatar
    Homework Helper

    ok well you know [itex]\Phi=BA[/itex]

    and Faraday's law is that the emf induced.[itex]E=-\frac{-d\Phi}{dt}[/itex]
    so

    [tex]E= - \frac{d\Phi}{dt} = - \frac{d}{dt}(BA)[/tex]

    B is constant so you can can remove it from inside the brackets...Can you see it better now?
     
  4. Mar 13, 2008 #3
    So it's like this:

    [tex]\Phi = BA[/tex]

    [tex]- \frac{d\Phi }{dt} =-B \frac{dA}{dt}[/tex]

    [tex]E = -B(-c) = Bc[/tex]

    which implies that B = E/c (assuming I didnt make any mistakes), but it says I need to express it in terms of A, c, or V.

    Did I make a mistake anywhere?
     
  5. Mar 13, 2008 #4

    rock.freak667

    User Avatar
    Homework Helper

    Well E is really V..so V=Bc.
     
  6. Mar 13, 2008 #5
    So for the second part of the question, I have to find the value of c and terms of v and L

    [tex]c = \frac{V}{B} = \frac{vLB}{B} = vL[/tex]

    I didnt make any stupid mistakes, did I?
     
  7. Mar 13, 2008 #6

    rock.freak667

    User Avatar
    Homework Helper

    Or you could have done it in a different way and say that in 1s the coil moves vm so that the area swept out in 1s is vLm^2. meaning that dA/dt=-c=vL.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: A Simple Way to Measure Magnetic Fields
Loading...