Consider the vector field given by (µ_0 and I are constants):
B=(µ_0I/2π)((-yi+xj)/(x^2+y^2))
B is the magnetic field around a wire along the z-axis carrying a constant current I in the z-direction.
1. Ready:
(a) Determine B ∙dr on any radial line of the form y=mx, where m is a constant.
(b) Determine B ∙dr on any circle of the form x^2 +y^2 =a^2, where a is a constant.
2. Go: For each of the following curves Ci , evaluate the line integral ∫Ci B∙dr
(a) C1, the top half of the circle r =5, traversed in a counterclockwise direction.
(b) C2, the top half of the circle r =2, traversed in a counterclockwise direction.
(c) C3, the top half of the circle r= 2, traversed in a clockwise direction.
(d) C4, the bottom half of the circle r 2, traversed in a clockwise direction.
(e) C5, the radial line from (2,0) to (5,0).
(f) C6, the radial line from (-5,0) to (-2,0).
3. Food for Thought
(a) Find closed curves C7 and C8 such that this integral is nonzero over C7 and
zero over C8 . It is enough to draw your curves; you do not need to
parameterize them.
(b) Ampère’s Law says that, for any closed curve C, this integral is (µ_0 times) the
current flowing through C (in the z-direction). Can you use this fact to explain
your results to part (a)?
(c) B conservative? Explain.
B=(µ_0I/2π)((-yi+xj)/(x^2+y^2))
B is the magnetic field around a wire along the z-axis carrying a constant current I in the z-direction.
1. Ready:
(a) Determine B ∙dr on any radial line of the form y=mx, where m is a constant.
(b) Determine B ∙dr on any circle of the form x^2 +y^2 =a^2, where a is a constant.
2. Go: For each of the following curves Ci , evaluate the line integral ∫Ci B∙dr
(a) C1, the top half of the circle r =5, traversed in a counterclockwise direction.
(b) C2, the top half of the circle r =2, traversed in a counterclockwise direction.
(c) C3, the top half of the circle r= 2, traversed in a clockwise direction.
(d) C4, the bottom half of the circle r 2, traversed in a clockwise direction.
(e) C5, the radial line from (2,0) to (5,0).
(f) C6, the radial line from (-5,0) to (-2,0).
3. Food for Thought
(a) Find closed curves C7 and C8 such that this integral is nonzero over C7 and
zero over C8 . It is enough to draw your curves; you do not need to
parameterize them.
(b) Ampère’s Law says that, for any closed curve C, this integral is (µ_0 times) the
current flowing through C (in the z-direction). Can you use this fact to explain
your results to part (a)?
(c) B conservative? Explain.
-
This is one long question...
1) Note that B ∙ dr = (µ₀ I/(2π)) (-y dx + x dy)/(x^2 + y^2)
a) Since y = mx,
B ∙ dr
= (µ₀ I/(2π)) (-mx dx + x * m dx)/(x^2 + (mx)^2)
= 0.
b) Using x = a cos t, y = a sin t,
B ∙ dr
= (µ₀ I/(2π)) [-(a sin t) (-a sin t dt) + (a cos t)(a cos t dt)]/[a^2 cos^2(t) + a^2 sin^2(t)]
= (µ₀ I/(2π)) a^2 dt / a^2
= µ₀ I dt / (2π).
-----------------
2) Use the results from #1 to do this efficiently.
a,b) ∫ci B ∙ dr = ∫(t = 0 to π) µ₀ I dt / (2π) = µ₀ I/2.
c) Use a negative sign to change orientation ==> -µ₀ I/2.
d) As with part c for the signage: -∫(t = π to 2π) µ₀ I dt / (2π) = -µ₀ I/2.
e,f) Here, m = 0 ==> ∫ci B ∙ dr = ∫ci 0 dt = 0.
----------------
3) For C₇, use a half circle arc (counterclockwise) with the x-axis 'diameter' between
(-a, 0) and (a, 0) to close off the arc. By the results in 2, this equals µ₀ I/2 + 0 = µ₀ I/2.
Let C₈ = C₁ U C₃ U C₅ U C₆, and use the results from 2 for each of the individual paths.
Since we get different answers for the integral on different closed paths, the integral most certainly is not independent of path. Observe how one contour avoids the origin, while the other does not.
I hope this helps!
1) Note that B ∙ dr = (µ₀ I/(2π)) (-y dx + x dy)/(x^2 + y^2)
a) Since y = mx,
B ∙ dr
= (µ₀ I/(2π)) (-mx dx + x * m dx)/(x^2 + (mx)^2)
= 0.
b) Using x = a cos t, y = a sin t,
B ∙ dr
= (µ₀ I/(2π)) [-(a sin t) (-a sin t dt) + (a cos t)(a cos t dt)]/[a^2 cos^2(t) + a^2 sin^2(t)]
= (µ₀ I/(2π)) a^2 dt / a^2
= µ₀ I dt / (2π).
-----------------
2) Use the results from #1 to do this efficiently.
a,b) ∫ci B ∙ dr = ∫(t = 0 to π) µ₀ I dt / (2π) = µ₀ I/2.
c) Use a negative sign to change orientation ==> -µ₀ I/2.
d) As with part c for the signage: -∫(t = π to 2π) µ₀ I dt / (2π) = -µ₀ I/2.
e,f) Here, m = 0 ==> ∫ci B ∙ dr = ∫ci 0 dt = 0.
----------------
3) For C₇, use a half circle arc (counterclockwise) with the x-axis 'diameter' between
(-a, 0) and (a, 0) to close off the arc. By the results in 2, this equals µ₀ I/2 + 0 = µ₀ I/2.
Let C₈ = C₁ U C₃ U C₅ U C₆, and use the results from 2 for each of the individual paths.
Since we get different answers for the integral on different closed paths, the integral most certainly is not independent of path. Observe how one contour avoids the origin, while the other does not.
I hope this helps!