A disk rotating about a point on its circumference - 2

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Consider the above figure, which shows the charged disk rotating about the point $\displaystyle P$ on its circumference.

Let $\displaystyle \sigma = \frac{Q}{\pi R^2}$ be the surface charge density of the disk.

At a general angle $\displaystyle\theta$ , draw a partial ring of radius $\displaystyle R+x$ , and thickness $\displaystyle dx$ .

Let us find the charge on this part of the ring.

$dq = \sigma\times (R+x)(2\theta)\times dx$

When this partial ring rotates, it will behave as a full ring, which is of radius $\displaystyle R+x$ , and carries a current $\displaystyle di$

$di = \frac{dq}{T} = \frac{dq}{\frac{2\pi}{w}} = \frac{w\cdot dq}{2\pi}$

The magnetic field produced by this current-carrying ring is,

$dB = \frac{\mu_o\cdot di}{2(R+x)}$

Substituting the value of $\displaystyle di$ ,

$dB = \frac{\mu_o}{2(R+x)} \frac{w\cdot dq}{2\pi}$

$\Rightarrow dB = \frac{\mu_o}{2(R+x)} \frac{w}{2\pi} \sigma (R+x) (2\theta) dx$

$dB = \frac{\mu_o}{2(R+x)} \frac{w}{2\pi} \frac{Q}{\pi R^2} (R+x)(2\theta) dx$

$dB = \frac{\mu_o w Q \theta}{2\pi^2R^2} dx$

Now, to get $\displaystyle x$ in terms of $\displaystyle \theta$ , it is just a matter of simple geometry.

Hint: Use the Sine rule

$\frac{\sin\theta}{R} = \frac{\sin(\pi-2\theta)}{R+x}$

$\Rightarrow x = R(2\cos\theta - 1)$

$\Rightarrow dx = R(-2\sin\theta d\theta)$

Substituting $\displaystyle dx$ in terms of $\displaystyle d\theta$ , and integrating the expression with the limits as : $\displaystyle \theta$ goes from $\displaystyle 0$ to $\displaystyle \frac{\pi}{2}$ , we get,

$B = \frac{\mu_o Qw}{\pi^2R}$

Therefore,

$\boxed{k = \frac{1}{\pi^2} = 0.10132}$

I have already posted the same diagram somewhere else (:P)

Consider a small element (an arc , shown in red, with radius $x$ ) having charge $dq = \sigma \times 2 \theta x dx$ . The effective current due to the rotation of this element is $di = dq \dfrac{\omega}{2 \pi}$

Now, using cosine rule in $\triangle OPQ,$

$\cos \theta = \dfrac{x^2+R^2-R^2}{2xR} = \dfrac{x}{2R}$

Now, $dB = \dfrac{\mu_{0} di}{2 x} = \dfrac{\mu_{0} dq \omega}{4 \pi x}$

= $\dfrac{\mu_{0} \sigma \omega}{2 \pi} \theta dx$

$\Rightarrow B = \displaystyle \int dB = \dfrac{\mu_{0} \sigma \omega}{2 \pi} \displaystyle \int_{0}^{2R} \cos^{-1} \bigg(\dfrac{x}{2R}\bigg) dx$

= $\dfrac{\mu_{0} \sigma \omega R}{\pi} \displaystyle \int_{0}^{1} \cos^{-1} z dz$

= $\dfrac{\mu_{0} \sigma \omega R}{ \pi} = \dfrac{\mu_{0} Q \omega}{\pi^2 R}$

Hence, $k = \frac{1}{\pi^2} = 0.101$