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A Simple Way to Measure Magnetic Fields

  • Thread starter cse63146
  • Start date
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1. Homework Statement

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.

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2. Homework 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:

Answers and Replies

rock.freak667
Homework Helper
6,230
31
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?
 
452
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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?
 
rock.freak667
Homework Helper
6,230
31
Well E is really V..so V=Bc.
 
452
0
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?
 
rock.freak667
Homework Helper
6,230
31
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.
 

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