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 + z²)¹/² For p large, B = curl(A), where 1/2 ^-(-3-30) A = b (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? O S A(c(t)) = =(- A(c(t)) = So, A(c(t)) = c(1). Therefore, A is parallel to c'(1) and tangent to C. R cos(t) R sin(t) (- A(c(t)) = A(c(t)) (0,0) So, A(c(1)) = -c'(1). Therefore, A is parallel to c'(t) and tangent to C. R sin(t) R cos(1) (sin() cos(0,0) and c'(t) = (R cos(t), - R sin(t), 0) O . BdS = R sin(t) R cos(1) 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(t), R cos(t), 0) So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. and c'(t) = (-R sin(t), R cos(t), 0) = Rin(0,0) and c'(t) = (R cos(1), - R sin(t), 0) (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.)
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 + z²)¹/² For p large, B = curl(A), where 1/2 ^-(-3-30) A = b (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? O S A(c(t)) = =(- A(c(t)) = So, A(c(t)) = c(1). Therefore, A is parallel to c'(1) and tangent to C. R cos(t) R sin(t) (- A(c(t)) = A(c(t)) (0,0) So, A(c(1)) = -c'(1). Therefore, A is parallel to c'(t) and tangent to C. R sin(t) R cos(1) (sin() cos(0,0) and c'(t) = (R cos(t), - R sin(t), 0) O . BdS = R sin(t) R cos(1) 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(t), R cos(t), 0) So, A(c(t)) c'(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C. and c'(t) = (-R sin(t), R cos(t), 0) = Rin(0,0) and c'(t) = (R cos(1), - R sin(t), 0) (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.)
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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 + z²)¹/² For p large, B = curl(A), where
A = (-33, -3,0)
R
Current loop
(a) Let C be a horizontal circle of radius R with center (0, 0, c), and parameterization c(t) where c is large.
Which of the following correctly explains why A is tangent to C?
A(c(t)) =
So, A(c(t)) =
A(c(t))
-(-²
A(c(t)) =
=
A(c(t))
cos(0,0)
p3
(1). Therefore, A is parallel to c'(t) and tangent to C.
Rcos(t) R sin(t)
= (-OS
R sin(t) R cos(1)
p³
So, A(c(1)) = -c'(1). Therefore, A is parallel to c'(t) and tangent to C.
O
BdS =
=
and c'(t)= (-R sin(t), R cos(t), 0)
Rin(1,0) and c'(t) = (R cos(1), -R sin(1), 0)
So, A(c(t)) c(t) = 0. Therefore, A is perpendicular to c'(t) and tangent to C.
O
R sin(1) R cos(1)
(R$ R COS(0,0) and c'(t) = (R cos(t), - R sin(t), 0)
R cos(1)
p3
So, A(c(t)) - c'(t) = 0. Therefore, A is perpendicular to c' (t) and tangent to C.
R sin(t)
-
and c'(t)= (-R sin(t), R cos(t), 0)
(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.)
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