Rotation about edge!

Electricity and Magnetism Level 5

A uniformly Charged Disc is rotating with constant angular velocity ω \omega about it's edge 'P' . Then Find Ratio

B p V p × 10 16 \displaystyle{\cfrac { { B }_{ p } }{ { V }_{ p } } \times { 10 }^{ 16 }} .

Here B p & V p { B }_{ p }{ \& V }_{ p }\quad means magnitude of Magnetic field strength and electric Potential at the edge P.

Details

∙ \bullet Assume that whole Experiment is doing in Vacuum.
∙ \bullet Use 1 μ o ϵ o = 3 × 10 8 \displaystyle{{ \cfrac { 1 }{ \sqrt { { \mu }_{ o }{ \epsilon }_{ o } } } =3\times { 10 }^{ 8 } }}

∙ \bullet use ω = 126 \omega =126\quad (All are in SI unit).

Try more Mixing of concepts


The answer is 14.

Deepanshu Gupta by Deepanshu Gupta , 25, India

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2 solutions

Consider a piece of charge at ( r , θ ) (r,\theta) and has charge dq

Then the Magnetic field is d B ⃗ = μ 0 4 π d q × ω × r r 2 z ^ d \vec{B}=\dfrac{\mu_{0}}{4\pi} \dfrac{dq\times \omega \times r}{r^{2}} \hat{z}

And the Electric voltage is d V = 1 4 π ϵ 0 d q r dV=\dfrac{1}{4\pi \epsilon_{0}} \dfrac{dq}{r}

So, we can see that ∣ d B ⃗ ∣ d V = μ 0 ϵ 0 ω = c o n s t . \dfrac{\left| d \vec{B} \right| }{dV}=\mu_{0} \epsilon_{0} \omega=const.

Hence it's true for all ratio of B and V.

Then ∣ B ⃗ ∣ V × 1 0 16 = 126 × 1 0 16 9 × 1 0 16 = 14 \dfrac{\left| \vec{B} \right| }{V} \times 10^{16}=\dfrac{126 \times 10^{16}}{9 \times 10^{16}}=\boxed{14}

4 Helpful 0 Interesting 2 Brilliant 0 Confused
Jatin Yadav
Jan 22, 2015

A disk rotating about a point on its circumference-2 and Potential with an edge are already there

5 Helpful 0 Interesting 0 Brilliant 0 Confused

Before someone writes a solution to this involving figures and arcs of circles, let me say something. Everybody used cosine formula to establish relationship between R R and x x (to know what I'm talking about, refer to the questions suggested by Jatin Yadav). Why cosine formula? Isn't it obvious that x = 2 R c o s θ x=2Rcos\theta since the ends of the diameter and side x x form a right triangle(angles in semicircle are π / 2 \pi/2 )? I know it doesn't make much difference, but still...

Raghav Vaidyanathan - 6 years, 4 months ago

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I have a better way to calculate magnetic field. Just cut arcs from that point and (we can derive magnetic field due to charged rind and hence an arc rotating at some w) so integrate it over x where x is radii of the arc considered

Jatin Narde - 5 years, 10 months ago

Already there? I did nt get u.

Jatin Narde - 5 years, 10 months ago

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Sorry, I m new, so I did nt knew that u r referring to a problem.

Jatin Narde - 5 years, 10 months ago

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