# Integral of an integral of a dot product?

• 2thumbsGuy
In summary, the induced emf in a coil with N loops and radius R surrounding a long solenoid with radius r and n turns per meter, where the solenoid has a sinusoidal current with maximum value I0 and frequency f, can be calculated using the equation ε=-2π2μ0fNnr2I0cos(2πft). This is derived using the equations I(t)=I0sin(2πft), B=\mu0nI, and ε=N|(d\PhiB)/dt|, where the dot product between the magnetic field vector and the area vector is taken and the integral is evaluated over the area of the coil. The negative sign in the final equation is due to the
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!

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

Last edited:
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
$$\vec{dA} = \begin{pmatrix} 0 \\ 0 \\ dA \end{pmatrix} = \begin{pmatrix} 0 \\ 0 \\ dxdy \end{pmatrix}$$

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

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:
$$\int \int \mu_0 n I_0 Sin(2 \pi ft) dxdy = \mu_0 n I_0 Sin(2 \pi ft) \int \int dxdy = \mu_0 n I_0 Sin(2 \pi ft) A$$

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.

## 1. What is an integral of an integral of a dot product?

An integral of an integral of a dot product is a mathematical operation that involves taking the integral (or area under the curve) of a function that is itself an integral of a dot product of two vectors.

## 2. What is the purpose of calculating an integral of an integral of a dot product?

The purpose of calculating an integral of an integral of a dot product is to find the overall area under the curve of a function that is made up of multiple dot products. This can be useful in applications such as calculating work done by a varying force over a distance.

## 3. How do you solve an integral of an integral of a dot product?

To solve an integral of an integral of a dot product, you must first evaluate the innermost integral (dot product) and then integrate the resulting function. This can be done either symbolically or numerically using techniques such as substitution or integration by parts.

## 4. Can an integral of an integral of a dot product be negative?

Yes, an integral of an integral of a dot product can be negative. This would occur if the function being integrated has areas above and below the x-axis, resulting in positive and negative areas cancelling each other out.

## 5. What are some real-world applications of an integral of an integral of a dot product?

An integral of an integral of a dot product can be used in physics to calculate work done by varying forces, in economics to calculate the area under a demand curve, and in engineering to calculate the amount of energy used over a period of time. It is also commonly used in calculus and multivariable calculus courses to demonstrate the concept of multiple integrals.

Replies
9
Views
1K
Replies
9
Views
1K
Replies
9
Views
674
Replies
4
Views
663
Replies
4
Views
1K
Replies
5
Views
1K
Replies
3
Views
1K
Replies
19
Views
2K
Replies
3
Views
584
Replies
4
Views
1K