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Dipole and a detector

  • #1
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1. Homework Statement
A detector hjaving a radius 10 cm is placed 1 m away from an electric dipole. What fraction of total solid angle does the detector cover and what fraction of total dipole power does it detect (assuming 100% detection efficiency)

2. The attempt at a solution
I drew the diagram as well maybe that will help. In any case, i think the detector will make a cone (?) with the dipole.Since the base of this triangle has radius of 0.1 m and the height of the triangle is 1m, the apex angle is 5.7 degrees or [itex] 57\pi/180 [/itex]

the total angle swept by the triangle is thus
[tex] \int_{0}^{2\pi}\int_{0}^{5.7\pi/180}\sin\theta\ d \theta d\phi = \frac{5.7\pi}{90}[/tex]

the total angle swept by a sphere is 4pi so the fraction is simply the answer above divided by 4pi. Is that correct?
 

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Answers and Replies

  • #2
StatusX
Homework Helper
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You set up the integral correctly, but I get (I find it nice to work exactly when possible, although you'll have to round eventually anyway, so this isn't really that worth it)

[tex] \int_{0}^{2\pi}\int_{0}^{\theta_o}\sin\theta\ d \theta d\phi = 2 \pi \left( -\cos \theta |_0^{\theta_o} \right) = 2\pi(1 - \cos\theta_o) [/tex]

Or using [itex]\theta_o = \tan^{-1} (0.1)[/itex]:

[tex] 0.1 = \tan \theta_o = \frac{\sin \theta_o}{\cos \theta_o} = \frac{\sqrt{1-\cos^2 \theta_o}}{\cos \theta_o}[/tex]

[tex] \Rightarrow \cos \theta_o = \frac{1}{\sqrt{ (0.1)^2 + 1 } } [/tex]

So the integral becomes:

[tex] 2\pi \left(1 - \frac{1}{\sqrt{ (0.1)^2 + 1 } } \right) \approx 0.00993 \pi [/tex]

For a solid angle of (dividing by [itex] 4\pi[/itex], as you say) .00248


Usually, though, it suffices to approximate the solid angle by taking the area of object and dividing by [itex]4\pi r^2[/itex], where r is its distance from the dipole. This ignores the curvature of the sphere, and so is not exact, but is usually a good approximation. In this case, you would get [itex]\pi(0.1)^2/4 \pi (1)^2 = 0.0025[/itex], which is slightly off, but not too bad for most purposes.
 
Last edited:
  • #3
1,444
2
You set up the integral correctly, but I get (I find it nice to work exactly when possible, although you'll have to round eventually anyway, so this isn't really that worth it)

[tex] \int_{0}^{2\pi}\int_{0}^{\theta_o}\sin\theta\ d \theta d\phi = 2 \pi \left( -\cos \theta |_0^{\theta_o} \right) = 2\pi(1 - \cos\theta_o) [/tex]

Or using [itex]\theta_o = \tan^{-1} (0.1)[/itex]:

[tex] 0.1 = \tan \theta_o = \frac{\sin \theta_o}{\cos \theta_o} = \frac{\sqrt{1-\cos^2 \theta_o}}{\cos \theta_o}[/tex]

[tex] \Rightarrow \cos \theta_o = \frac{1}{\sqrt{ (0.1)^2 + 1 } } [/tex]

So the integral becomes:

[tex] 2\pi \left(1 - \frac{1}{\sqrt{ (0.1)^2 + 1 } } \right) \approx 0.00993 \pi [/tex]

For a solid angle of (dividing by [itex] 4\pi[/itex], as you say) .00248


Usually, though, it suffices to approximate the solid angle by taking the area of object and dividing by [itex]4\pi r^2[/itex], where r is its distance from the dipole. This ignores the curvature of the sphere, and so is not exact, but is usually a good approximation. In this case, you would get [itex]\pi(0.1)^2/4 \pi (1)^2 = 0.0025[/itex], which is slightly off, but not too bad for most purposes.
Thanks a lot

Ok now for the fraction of total dipole power detected (assuming 100% detection efficiency)

the total dipole power detected is given by [tex] \frac{dP}{d\Omega}=\frac{q^2 a^2}{2\pi c^3}\sin^2\theta[/tex]

So i need to integrate the above expression w.r.t. omega?

[tex] P = \frac{q^2 a^2}{4\pi c^3} \int_{0}^{2\pi}\int^{\theta_{0}}_{0}\sin^3\theta d\theta d\phi [/tex]

here theta zero is the angle calculated in the first section??

Is that correct?

I am working in CGS units so the formulas will look like their SI counterparts (obviously)
 

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