Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
The magnetic field B due to a small current loop (which is placed at the origin) is called a magnetic dipole, Let p = (x² + y² +₂²) 1/2 For large, B P = curl(A), where ^= (-3.₂0) A $ Current loop (a) Let C be a horizontal circle of radius R with center (0, 0, c), and parameterization c(1) where c is large. O Which of the following correctly explains why A is tangent to C? =(-Rsi A(c(t)) = COS(1), 0) and c'(t)= (-R sin(t), R cos(1), 0) So, A(c(1)) = c(1). Therefore, A is parallel to c'() and tangent to C. A(c(t)) = R sin(1) R cos(1) A(c(t)) = ) = (-R So, A(c(1)) = -c'(1). Therefore, A is parallel to c'(t) and tangent to C. R sin(t) R cos(t) =(. and c'(t) = (R cos(t), -R sin(t), 0) 3 So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. R cos(1) R sin(1) and c'(1) = (R cos(1), - R sin(t), 0)

The magnetic field B due to a small current loop (which is placed at the origin) is called a magnetic dipole, Let p = (x² + y² +₂²) 1/2 For large, B P = curl(A), where ^= (-3.₂0) A $ Current loop (a) Let C be a horizontal circle of radius R with center (0, 0, c), and parameterization c(1) where c is large. O Which of the following correctly explains why A is tangent to C? =(-Rsi A(c(t)) = COS(1), 0) and c'(t)= (-R sin(t), R cos(1), 0) So, A(c(1)) = c(1). Therefore, A is parallel to c'() and tangent to C. A(c(t)) = R sin(1) R cos(1) A(c(t)) = ) = (-R So, A(c(1)) = -c'(1). Therefore, A is parallel to c'(t) and tangent to C. R sin(t) R cos(t) =(. and c'(t) = (R cos(t), -R sin(t), 0) 3 So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. R cos(1) R sin(1) and c'(1) = (R cos(1), - R sin(t), 0)

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
See similar textbooks
icon
Related questions
Question
The magnetic field B due to a small current loop (which is placed at the origin) is called a magnetic dipole. Let p = (x² + y² +2²) ¹/2 For p large, B - curl(A), where ^= (-3, -3,0) (a) Let C be a horizontal circle of radius R with center (0, 0, c), and parameterization c(t) where c is large. Current loop Which of the following correctly explains why A is tangent to C? A(c(1)) = (-Rsin(1) -(-Rsin(1) So, A(c(t)) = c(1). Therefore, A is parallel to c'() and tangent to C. O )= (-²₁ ISB A(c(t)) = BdS= So, A(c(1)) = -c'(1). Therefore, A is parallel to c'(1) and tangent to C. A(c(t)) = cos(1) CO(0) (= (Rin A(c(1)) = = R sin(t) R cos(1) p3 So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. O R cos(1) R sin(t) and c'(t) = (-R sin(1), R cos(1), 0) R cos(1) R sin(t) R sin(0,0) and c'(t) = (R cos(1), - R sin(1), 0) Incorrect (,0). and c'(t) = (R cos(t), -R sin(1), 0) So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. (b) Suppose that R = 6 and c = 900. Use Stokes' Theorem to calculate the flux of B through C. (Use decimal notation. Round your answer to eight decimal places.) and c'(t)= (-R sin(t), R cos(1), 0)
Transcribed Image Text:The magnetic field B due to a small current loop (which is placed at the origin) is called a magnetic dipole. Let p = (x² + y² +2²) ¹/2 For p large, B - curl(A), where ^= (-3, -3,0) (a) Let C be a horizontal circle of radius R with center (0, 0, c), and parameterization c(t) where c is large. Current loop Which of the following correctly explains why A is tangent to C? A(c(1)) = (-Rsin(1) -(-Rsin(1) So, A(c(t)) = c(1). Therefore, A is parallel to c'() and tangent to C. O )= (-²₁ ISB A(c(t)) = BdS= So, A(c(1)) = -c'(1). Therefore, A is parallel to c'(1) and tangent to C. A(c(t)) = cos(1) CO(0) (= (Rin A(c(1)) = = R sin(t) R cos(1) p3 So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. O R cos(1) R sin(t) and c'(t) = (-R sin(1), R cos(1), 0) R cos(1) R sin(t) R sin(0,0) and c'(t) = (R cos(1), - R sin(1), 0) Incorrect (,0). and c'(t) = (R cos(t), -R sin(1), 0) So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. (b) Suppose that R = 6 and c = 900. Use Stokes' Theorem to calculate the flux of B through C. (Use decimal notation. Round your answer to eight decimal places.) and c'(t)= (-R sin(t), R cos(1), 0)
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 4 steps with 2 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Recommended textbooks for you
Advanced Engineering Mathematics
Advanced Engineering Mathematics
Advanced Math
ISBN:
9780470458365
Author:
Erwin Kreyszig
Publisher:
Wiley, John & Sons, Incorporated
Numerical Methods for Engineers
Numerical Methods for Engineers
Advanced Math
ISBN:
9780073397924
Author:
Steven C. Chapra Dr., Raymond P. Canale
Publisher:
McGraw-Hill Education
Introductory Mathematics for Engineering Applicat…
Introductory Mathematics for Engineering Applicat…
Advanced Math
ISBN:
9781118141809
Author:
Nathan Klingbeil
Publisher:
WILEY
Mathematics For Machine Technology
Mathematics For Machine Technology
Advanced Math
ISBN:
9781337798310
Author:
Peterson, John.
Publisher:
Cengage Learning,
Basic Technical Mathematics
Basic Technical Mathematics
Advanced Math
ISBN:
9780134437705
Author:
Washington
Publisher:
PEARSON
Topology
Topology
Advanced Math
ISBN:
9780134689517
Author:
Munkres, James R.
Publisher:
Pearson,
SEE MORE TEXTBOOKS