Ampere's Law?

Electricity and Magnetism Level 5

An infinitely long hollow circular tube of radius R R carries a non-uniformly distributed current (into the page). The infinitesimal current associated with a particular angle θ \theta is:

d I = s i n θ d θ dI = sin \theta \, d \theta

Thus, the total current enclosed by an "Amperian" loop concentric with the tube is zero. The magnitude of the magnetic flux density ( B ) (B) at a test point located a distance 2 R 2 R from the center of the tube can be expressed as:

B = α μ 0 R B = \alpha \, \frac{\mu_0}{R}

What is the value of α \alpha ?

Note: The expression given for B B only relates magnitudes, and is not concerned with units.


The answer is 0.125.

Steven Chase by Steven Chase , 37, USA

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Laszlo Mihaly
May 7, 2018

The magnetic field of a straight wire carrying a current I I at a distance r r is μ 0 I 2 π r \frac{\mu_0 I}{2 \pi r} . A d θ d\theta element of the cylinder will contribute d B = μ 0 sin ⁡ θ d θ 2 π r dB= \frac{\mu_0\sin \theta d\theta}{2 \pi r} , where the green line in the Figure represents r r . The magnetic field is perpendicular to the green line and therefore it makes an angle β \beta to the vertical . Due to the opposite directions of the currents in the upper and lower half of the cylinder, the vertical component of the magnetic fields will cancel. The horizontal component is d B x = μ 0 2 π sin ⁡ θ sin ⁡ β d θ r dB_x=\frac{\mu_0}{2 \pi}\frac{\sin \theta \sin \beta d\theta}{r} . Simple geometry tells us that r sin ⁡ θ = R sin ⁡ β r \sin \theta=R \sin \beta and r 2 = x 2 + R 2 − 2 x R cos ⁡ θ = R 2 ( 5 − 4 cos ⁡ θ ) r^2=x^2+R^2-2xR \cos \theta=R^2 (5-4\cos \theta) , where x = 2 R x=2R for our selected point. All of this together yields

B = μ 0 2 π ∫ 0 2 π R sin ⁡ 2 θ r 2 d θ = μ 0 2 π R ∫ 0 2 π sin ⁡ 2 θ 5 − 4 cos ⁡ θ d θ B= \frac {\mu_0}{2\pi} \int_{0}^{2\pi} R \frac {\sin^2\theta}{r^2}d\theta = \frac {\mu_0}{2\pi R} \int_{0}^{2\pi} \frac {\sin^2\theta}{5-4\cos\theta} d\theta

The integral can be evaluated (e.g. with Wolfram alpha) and it yields π / 4 \pi/4 . Therefore the parameter in question is α = 1 / 8 = 0.125 \alpha=1/8=0.125 .

4 Helpful 0 Interesting 1 Brilliant 0 Confused

And presumably, if we calculated the vector flux density at all points on an Amperian loop concentric with the tube, and took the loop integral, that would be zero, since the enclosed current is zero. I didn't actually verify that, but it's a sure bet.

Steven Chase - 3 years, 1 month ago

Log in to reply

I agree, it must be zero.

Laszlo Mihaly - 3 years, 1 month ago

To integrate it by hand, change the integration interval to [ − π , π ] [-\pi,\:\pi] (possible, because the integrand is periodic). Now we may use general trig- or Weierstrass-Substitution t : = tan ⁡ ( θ 2 ) ( ∗ ) t:=\tan(\frac{\theta}{2})\quad (*) and partial fraction decomposition (PFD): ∫ − π π s θ 2 5 − 4 c θ d θ = ( ∗ ) ∫ R 4 t 2 [ 5 ( 1 + t 2 ) − 4 ( 1 − t 2 ) ] ( 1 + t 2 ) ⋅ 2 d t 1 + t 2 = 8 9 ∫ R t 2 d t ( t 2 + 1 9 ) ( t 2 + 1 ) 2 = PFD 1 8 ∫ R − 1 t 2 + 1 9 + 8 ( t 2 + 1 ) 2 + 1 t 2 + 1 d t = 1 8 [ − 3 arctg ⁡ ( 3 t ) + 4 ( arctg ⁡ ( t ) + t 1 + t 2 ) + arctg ⁡ ( t ) ] − ∞ ∞ = π 4 \begin{aligned} \int_{-\pi}^\pi\frac{s^2_\theta}{5-4c_\theta}\:d\theta & \underset{(*)}{=}\int_{\mathbb{R}}\frac{4t^2}{ \left[5(1+t^2)-4(1-t^2)\right](1+t^2) }\cdot\frac{2\:dt}{1+t^2}=\frac{8}{9}\int_{\mathbb{R}}\frac{t^2\:dt}{\left(t^2+\frac{1}{9}\right)(t^2+1)^2}\\[.5em] &\underset{\text{PFD}}{=} \frac{1}{8}\int_{\mathbb{R}}\frac{-1}{t^2+\frac{1}{9}} + \frac{8}{(t^2+1)^2} + \frac{1}{t^2+1}\:dt=\frac{1}{8}\left[ -3\arctg(3t) + 4\left(\arctg(t) +\frac{t}{1+t^2}\right) +\arctg(t) \right]_{-\infty}^\infty=\frac{\pi}{4} \end{aligned}

Carsten Meyer - 2 weeks, 2 days ago

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...