Integral of an integral of a dot product?

[2thumbsGuy]

Homework Statement

A coil with N loops and radius R is surrounds a very long solenoid of radius r and n turns per meter.

The current in the solenoid is varying sinusoidally with time according to the relation I(t)=I₀sin(2πft)

where I₀ is the maximum value of the current, and f is its frequency. (This is the type of current supplied by the power company.) If R >> r, show that the induced emf in the coil is given by

ε=-2π²μ₀fNnr²I₀cos(2πft)

Homework Equations

I(t)=I₀sin(2πft)
$\Phi$B = ∫$\vec{B}$∙d$\vec{A}$
B = $\mu$₀nI
ε=N|(d$\Phi$B)$/$dt|

The Attempt at a Solution

Substituting the above,
ε=N|(d∫$\mu$₀nI₀sin(2πft)∙d$\vec{A}$)$/$dt|

ε=N$\mu$₀nI₀|(d∫sin(2πft)∙d$\vec{A}$)$/$dt|

∫sin(2πft) = -cos(2πft)/2π

I believe (d∫sin(2πft)∙d$\vec{A}$)$/$dt is a double integral, but if I evaluate I just end up with stuff in the denominator, which does not match with the answer.

I have no intuition for dot products and only a passable understanding of integrals, mathematically. I understand what they do, what they are for in layman's terms, but expressing it in mathematical terms is my perennial weakness. For the love of Congressman Weiner, help me to understand!

 

[lanedance]

hey 2 thumbs guy you can write a whole equation in tex eg.
$$\Phi_B = \int \vec{B} \bullet \vec{dA}$$

 

[lanedance]

note that B is a vector, so has a direction along the axis of the cylinder, call it the z direction
$$\vec{B} = \begin{pmatrix} 0 \\ 0 \\ \mu_0 n I \end{pmatrix}$$

similarly the element of area has a direction normal to the surface, as you are considering a cross section of the sphere, this will also be in the z direction, and can consider a scalar area element dA = dxdy
$$\vec{dA} = \begin{pmatrix} 0 \\ 0 \\ dA \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ dxdy \end{pmatrix}$$

 

[lanedance]

then the integral becomes
$$\Phi_B = \int \vec{B} \bullet \vec{dA} <br /> = \int \mu_0 n I_0 Sin(2 \pi ft) dA= \int \int \mu_0 n I_0 Sin(2 \pi ft) dxdy$$

Note the integral is over x & y not t, in fact the integrand doesn't depend on x or y, so just evaluates to the area of the circle, (A), multiplied by the integrand:
$$\int \int \mu_0 n I_0 Sin(2 \pi ft) dxdy = \mu_0 n I_0 Sin(2 \pi ft) \int \int dxdy = \mu_0 n I_0 Sin(2 \pi ft) A$$

 

[2thumbsGuy]

Thanks for the tex tips, lane, and the re-introduction of the dot product. But I guess I'm confused on why the B vector and dxdy are both on the z axis. Shouldn't they be perpendicular? I thought the area vector pointed normally from the axis, hence our concern for sin, and ultimately -cos.

And what you suggest mean that $$\mu_0 n I_0 Sin(2πft) A = πr^2 μ_0 nI_0 Sin(2πft)$$, which doesn't give me a negative cosine as I would expect.