Chapter 6: Q14P (page 335)

$∮ (yi - xj + zk) - dr$ around the circumference of the circle of radius $2$ , center at the origin, in the $(x, y)$ plane.

Short Answer

Expert verified

The solution derived is $I = - 8 π .$

Step by step solution

Given Information.

The given expression is $∮ (yi - xj + zk) - dr$

Definition of vector.

A quantity that has magnitude as well as direction is called a vector. It is typically denoted by an arrow in which the head determines the direction of the vector and the length determines its magnitude.

Apply Stokes’s theorem.

Apply Stokes' theorem and the fact that $\nabla \times V = - 2 \hat{k}, Z = 0$ and then the question can be solved as shown below.

$I = - 2 \int_{σ} dσ$

$= 2$ (Area of circle)

$= - 2 (4) π$

$= - 8 π$

Hence, the solution derived is $I = - 8 π$ .

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$∮ (yi - xj + zk) - dr$ around the circumference of the circle of radius $2$ , center at the origin, in the $(x, y)$ plane.

Short Answer

Step by step solution

Given Information.

Definition of vector.

Apply Stokes’s theorem.

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∮(yi−xj+zk)−draround the circumference of the circle of radius 2, center at the origin, in the (x,y)plane.

Short Answer

Step by step solution

Given Information.

Definition of vector.

Apply Stokes’s theorem.

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Linear Momentum

Atoms and Radioactivity

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Study anywhere. Anytime. Across all devices.

$∮ (yi - xj + zk) - dr$ around the circumference of the circle of radius $2$ , center at the origin, in the $(x, y)$ plane.