(i)Show that the magnetic field [latex]\mathbf{B}[/latex] on the axis

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SUMMARY

The magnetic field on the axis of a circular current loop of radius a is given by the formula \mathbf{B}=\frac{\mu_0 I a^2}{2(a^2+z^2)^{\frac{3}{2}}} \mathbf{\hat{z}}, where I is the current and z is the distance from the center of the loop. For an insulating disc with uniform surface charge density \sigma rotating at angular velocity \omega, the surface current density is defined as \mathbf{K(r)}=\sigma \mathbf{\omega} \wedge \mathbf{r}. The contribution to the magnetic field from a ring of radius r is expressed as B_z = \frac{\mu_0}{2} \int_{0}^{a} \frac{\sigma \omega r^3 dr}{(r^2+z^2)^{\frac{3}{2}}}. As z approaches infinity, the magnetic field behaves as \mathbf{B}(z) \sim \frac{1}{8} \mu_0 \sigma \omega \frac{a^4}{z^3} \mathbf{\hat{z}}.

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  • #31


Substitute that in your original expression and see. Incidentally you won't get back the original term ( the first expression you wrote) but the modified term for z tending to infinity. This is because you've calculated the field using that condition.
 
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  • #32


B(z)=\frac{1}{2} \mu_0 \sigma \omega \frac{r^3 dr}{z^3} as b-> a dr->0 though?
 
  • #33


That's what you should get.
 
  • #34


hmmm...surely if dr->0, then B(z)->0 which doesn't make much sense though?
 
  • #35


You should try to work out an expression for the current through this ring.
 

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