/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 28 A vector force field \(\mathbf{F... [FREE SOLUTION] | 91影视

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Chapter 11: Problem 28

A vector force field \(\mathbf{F}\) is defined in Cartesian coordinates by $$ \mathbf{F}=F_{0}\left[\left(\frac{y^{3}}{3 a^{3}}+\frac{y}{a} e^{x y / a^{2}}+1\right) \mathbf{i}+\left(\frac{x y^{2}}{a^{3}}+\frac{x+y}{a} e^{x y / a^{2}}\right) \mathbf{j}+\frac{z}{a} e^{x y / a^{2}} \mathbf{k}\right] $$ Use Stokes' theorem to calculate $$ \oint_{L} \mathbf{F} \cdot d \mathbf{r} $$ where \(L\) is the perimeter of the rectangle \(A B C D\) given by \(A=(0,1,0), B=(1,1,0)\), \(C=(1,3,0)\) and \(D=(0,3,0)\)

Short Answer

Expert verified
The circulation integral around rectangle ABCD is \( \frac{20 F_0}{a^3} \)

Step by step solution

01

Understand Stokes' Theorem

Stokes' theorem relates the circulation of a vector field \( \mathbf{F} \) around a closed loop L to the integral of the curl of \( \mathbf{F} \) over the surface S bounded by L. Mathematically, it's given by: \[ \oint_L \mathbf{F} \cdot d\mathbf{r} = \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} \]
02

Calculate the Curl of \( \mathbf{F} \)

To find the curl of \( \mathbf{F} \), use the formula: \[ abla \times \mathbf{F} = \left( \frac{\partial F_k}{\partial y} - \frac{\partial F_j}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_i}{\partial z} - \frac{\partial F_k}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_j}{\partial x} - \frac{\partial F_i}{\partial y} \right) \mathbf{k} \]. Insert the components of \( \mathbf{F} \) into this formula to get: \[ abla \times \mathbf{F} = F_0 \left[ 0 \mathbf{i} + 0 \mathbf{j} + \left( \frac{\partial}{\partial x} \left( \frac{x y^2}{a^3} + \frac{x+y}{a} e^{xy / a^2} \right) - \frac{\partial}{\partial y} \left( \frac{y^3}{3a^3} + \frac{y}{a} e^{xy / a^2} + 1 \right) \right) \mathbf{k} \right] \]
03

Compute the Partial Derivatives

Calculate each partial derivative separately: \[ \frac{\partial}{\partial x} \left( \frac{x y^2}{a^3} + \frac{x+y}{a} e^{xy / a^2} \right) = \frac{y^2}{a^3} + \frac{x y^2}{a^3} \left( \frac{y}{a^2} e^{xy / a^2} \right) + \frac{y}{a} e^{xy / a^2} = \frac{y^2}{a^3} + \frac{y^3}{a^5} e^{xy / a^2} + \frac{y}{a} e^{xy / a^2} \] and \[ \frac{\partial}{\partial y} \left( \frac{y^3}{3a^3} + \frac{y}{a} e^{xy / a^2} + 1 \right) = \frac{3y^2}{3a^3} + \left( \frac{x}{a} e^{xy / a^2} + \frac{y}{a} \left( \frac{x}{a^2} e^{xy / a^2} \right) \right) = \frac{y^2}{a^3} + \frac{x}{a} e^{xy / a^2} + \frac{xy}{a^3} e^{xy / a^2} \]
04

Simplify the Curl Expression

Combining the results from Step 3, the curl simplifies to: \[ abla \times \mathbf{F} = F_0 \left[ \left( \frac{y^2}{a^3} + \frac{y^3}{a^5} e^{xy / a^2} + \frac{y}{a} e^{xy / a^2} \right) - \left( \frac{y^2}{a^3} + \frac{x}{a} e^{xy / a^2} + \frac{xy}{a^3} e^{xy / a^2} \right) \right] \mathbf{k} \] which further simplifies to \[ abla \times \mathbf{F} = F_0 \left[ \frac{y^3 - xy^2 + y - xa e^{xy / a^2}}{a^3} \right] \mathbf{k} \]
05

Find the Surface Integral

Use the surface integral form of Stokes' theorem: \[ \iint_S (abla \times \mathbf{F}) \cdot d\mathbf{S} = \iint_S F_0 \left( \frac{y^2 (y - x) + y - x a e^{xy / a^2}}{a^3} \right) \mathbf{k} \cdot d\mathbf{S} \] Given the rectangle ABCD lies in the xy-plane with \( z = 0 \, d\mathbf{S} = dz dy \mathbf{k} \), and bounds y from 1 to 3, x from 0 to 1, this simplifies the integral: \[ \iint_S F_0 \left( \frac{y^2 (y - x) + y}{a^3} \right) dz dy \mathbf{k} \]. Evaluate this integral over the bounds.
06

Evaluate the Remaining Integral

The integral reduces to a double integral since \( z \) is 0: \[ F_0 \iint_S \left( \frac{y^2 (y - x) + y}{a^3} \right) dx dy \]. Integrating over the rectangle's parameters: \[ = F_0 \int_1^3 \int_0^1 \left( \frac{y^3 - xy^2 + y}{a^3} \right) dx dy \]
07

Solve the Final Integral

Integrating over x first: \[ \int_0^1 \left( \frac{y^3}{a^3} - \frac{xy^2}{a^3} + \frac{y}{a^3} \right) dx = \frac{y^3}{a^3}[1]_0^1 - \frac{y^2}{a^3} \left( \frac{x^2}{2} \right)[1]_0^1 + \frac{y}{a^3}[1]_0^1 = \frac{y^3 - \frac{y^2}{2} + y}{a^3} \] then, \[ F_0 \int_1^3 \left( \frac{y^3 - \frac{y^2}{2} + y}{a^3} \right) dy \]. Integrate over y: \[ = F_0 \left( \int_1^3 \frac{y^3}{a^3} dy - \frac{1}{2} \int_1^3 \frac{y^2}{a^3} dy + \int_1^3 \frac{y}{a^3} dy \right) = F_0 \left( \frac{1}{a^3} \left[ \frac{y^4}{4}\right]_1^3 - \frac{1}{2a^3} \left[ \frac{y^3}{3} \right]_1^3 + \frac{1}{a^3} \left[ \frac{y^2}{2} \right]_1^3 \right) \]
08

Evaluate the Definite Integrals

\[ = F_0 \left( \frac{1}{a^3} \left( \frac{81}{4} - \frac{1}{4} \right) - \frac{1}{2a^3} \left( 9 - \frac{1}{3} \right)+ \frac{1}{a^3} \left( \frac{9}{2} - \frac{1}{2} \right) \ = F_0 \left( \frac{80}{4a^3} - \frac{8}{a^3}+ \frac{8}{a^3} \right) = F_0 \left( \frac{20}{a^3} \right). \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Vector Force Field
A vector force field, represented as \(\text{{\mathbf{{F}}}}\), is a function that assigns a vector to each point in space. This vector indicates the force's direction and magnitude at that point. In the given exercise, \(\text{{\mathbf{{F}}}}\) is defined in Cartesian coordinates and has components along the \(\text{{\mathbf{{i}}}}, \text{{\mathbf{{j}}}}, \text{{\mathbf{{k}}}}\) directions, written as: \[ \mathbf{F}=F_{0}\left[\left(\frac{y^{3}}{3a^{3}}+\frac{y}{a}e^{x y/a^{2}}+1\right) \,\mathbf{i}+\left(\frac{x y^{2}}{a^{3}}+\frac{x+y}{a} e^{x y/a^{2}}\right) \,\mathbf{j}+\frac{z}{a} e^{x y/a^{2}} \, \mathbf{k}\right] \]. This means that at every point (x, y, z), the vectors indicate the force experienced.
Curl of a Vector Field
The curl of a vector field \[ \mathbf{F} \] measures the rotation or swirling strength of the field at a point. In mathematical terms, the curl of \(\text{\mathbf{F}}\) is given by the vector operator: \[ abla \times \mathbf{F} = \left( \frac{\partial F_{z}}{\partial y} - \frac{\partial F_{y}}{\partial z} \right) \mathbf{i} + \left( \frac{\partial F_{x}}{\partial z} - \frac{\partial F_{z}}{\partial x} \right) \mathbf{j} + \left( \frac{\partial F_{y}}{\partial x} - \frac{\partial F_{x}}{\partial y} \right) \mathbf{k} \].
Surface Integral
A surface integral is used to integrate over a surface in space. For the curl calculated earlier, we integrate it over the surface \(S\theta\) bounded by the loop \(L\theta\). In the context of Stokes' theorem, this surface integral is expressed as \[ \iint_{S} \left( abla \times \mathbf{F} \right) \cdot d\mathbf{S} \]. Here, d\(\mathbf{S} \) is the vector normal to the surface S.
Line Integral
A line integral calculates the cumulative effect of a vector field along a curve. In this exercise, we are required to evaluate the line integral around the closed loop \(\text{L}\theta\). Stokes' Theorem relates this line integral to the surface integral of the curl over the surface bounded by \(\text{L}\theta\): \(\text{\oint_{L} \mathbf{F} \cdot d\mathbf{r}} = \text{\iint_{S}}( abla \times \mathbf{F}) \cdot d\mathbf{S}\).

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Most popular questions from this chapter

\(\mathbf{F}\) is a vector field \(x y^{2} \mathbf{i}+2 \mathbf{j}+x \mathbf{k}\), and \(L\) is a path parameterised by \(x=c t, y=c / t\) \(z=d\) for the range \(1 \leq t \leq 2\). Evaluate (a) \(\int_{L} \mathbf{F} d t\), (b) \(\int_{L} \mathbf{F} d y\) and (c) \(\int_{L} \mathbf{F} \cdot d \mathbf{r}\).

A single-turn coil \(C\) of arbitrary shape is placed in a magnetic field \(\mathbf{B}\) and carries a current \(I .\) Show that the couple acting upon the coil can be written as $$ \mathbf{M}=I \int_{C}(\mathbf{B} \cdot \mathbf{r}) d \mathbf{r}-I \int_{C} \mathbf{B}(\mathbf{r} \cdot d \mathbf{r}) $$ For a planar rectangular coil of sides \(2 a\) and \(2 b\) placed with its plane vertical and at an angle \(\phi\) to a uniform horizontal field \(\mathbf{B}\), show that \(\mathbf{M}\) is, as expected, \(4 a b B I \cos \phi \mathbf{k}\)

One of Maxwell's electromagnetic equations states that all magnetic fields \(\mathbf{B}\) are solenoidal (i.e. \(\nabla \cdot \mathbf{B}=0\) ). Determine whether each of the following vectors could represent a real magnetic field; where it could, try to find a suitable vector potential \(\mathbf{A}\), i.e. such that \(\mathbf{B}=\nabla \times \mathbf{A}\). (Hint: seek a vector potential that is parallel to \(\nabla \times \mathbf{B} .\) : (a) \(\frac{B_{0} b}{r^{3}}\left[(x-y) z \mathbf{i}+(x-y) z \mathbf{j}+\left(x^{2}-y^{2}\right) \mathbf{k}\right]\) in Cartesians with \(r^{2}=x^{2}+y^{2}+z^{2}\) (b) \(\frac{B_{0} b^{3}}{r^{3}}\left[\cos \theta \cos \phi \hat{\mathbf{e}}_{r}-\sin \theta \cos \phi \hat{\mathbf{e}}_{\theta}+\sin 2 \theta \sin \phi \hat{\mathbf{e}}_{\phi}\right]\) in spherical polars; (c) \(B_{0} b^{2}\left[\frac{z \rho}{\left(b^{2}+z^{2}\right)^{2}} \hat{\mathbf{e}}_{\rho}+\frac{1}{b^{2}+z^{2}} \hat{\mathbf{e}}_{z}\right]\) in cylindrical polars.

The vector field \(\mathbf{F}\) is defined by $$ \mathbf{F}=2 x z \mathbf{i}+2 y z^{2} \mathbf{j}+\left(x^{2}+2 y^{2} z-1\right) \mathbf{k} $$ Calculate \(\nabla \times \mathbf{F}\) and deduce that \(\mathbf{F}\) can be written \(F=\nabla \phi\). Determine the form of \(\phi\).

Show that the expression below is equal to the solid angle subtended by a rectangular aperture, of sides \(2 a\) and \(2 b\), at a point on the normal through its centre, and at a distance \(c\) from the aperture: $$ \Omega=4 \int_{0}^{b} \frac{a c}{\left(y^{2}+c^{2}\right)\left(y^{2}+c^{2}+a^{2}\right)^{1 / 2}} d y $$ By setting \(y=\left(a^{2}+c^{2}\right)^{1 / 2} \tan \phi\), change this integral into the form $$ \int_{0}^{\phi_{1}} \frac{4 a c \cos \phi}{c^{2}+a^{2} \sin ^{2} \phi} d \phi $$ where \(\tan \phi_{1}=b /\left(a^{2}+c^{2}\right)^{1 / 2}\), and hence show that $$ \Omega=4 \tan ^{-1}\left[\frac{a b}{c\left(a^{2}+b^{2}+c^{2}\right)^{1 / 2}}\right] $$
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