Faraday's law lab report problem with equations and graph(slope)

[timsnewbridge]

Homework Statement

the problem is we drew up a graph o E^-3/2(y) against Z^2(x)
so the slope is [E^-3/2]/[Z^2]=m
we need to rewrite equation three given that z is not equal to zero
and then get a final equation like four.
we need to isolate (a) so we can determine a value for it using our slope value aswell
ie we need to get an eqaution a=( )
but i can't do this.

Homework Equations

(1)B=[u(N1)(i)(a^2)]/[2(a^2 + z^2)^(3/2)]

when z=0 B=[u(N1)(i)]/[2(a)]

B= magnetic field, N1= turns on large coil, N2=turns on pick-up coil
a=radius of large coil, z = dist from large coil, u= const=4(pi)(10^-7)
i= current through large coil= (Io)[sin(2(pi)(f)(t))] f=frequency, pi= 3.1415... t=250nano sec

(2)O=B(A2)(N2) O=the flux, A2=area of pickup coil, N2= number of turns on pick up coil

(3)E=[u(N1)(N2)(A2)(Io)(d/dt of sin(2(pi)(f)(t))]/[2a] when z=0
= [u(N1)(N2)(A2)(Io)(pi)(f)cos((2)(pi)(f)(t))]/[a] when z=0

The Attempt at a Solution

my attempted new version of the equation is as follows
E=[(a)^2(u)(N1)(N2)(A2)(Io)(Pi)(f)(cos((2)(pi)(f)(t)))]/[(a^2+z^2)^3/2]

but i can not isolate a, keeping in mind that [E^-3/2]/[Z^2]=m

please help, its more maths question :)

 

[AvroArrow]

I think the equation should be something like this:E=[(a)^2(u)(N1)(N2)(A2)(Io)(Pi)(f)(cos((2)(pi)(f)(t)))]/[(a^2+z^2)^3/2]Now, we can isolate a by taking the square root of both sides of the equation:a = sqrt([E^-3/2]/[Z^2])This gives us our final equation:a = sqrt([E^-3/2]/[Z^2])