Clarification about the problem "Danger? I laugh at danger!"

In this problem, when we consider rotation about the IAR, then it becomes as follows -

I1 I1

I2 I2

dB=μo4πdqdtdl×aa3=μo4πdq2vrsin(θ/2)d(r2+d2)3/2\begin{aligned} \mathrm{d}B &=\dfrac{\mu_o}{4\pi} \dfrac{\mathrm{d}q}{\mathrm{d}t} \dfrac{\vec{\mathrm{d}l} \times \vec{a}}{|\vec{a}|^3}\\ &=\dfrac{\mu_o}{4\pi}\mathrm{d}q \dfrac{2vr\sin(\theta/2) d}{(r^2+d^2)^{3/2}} \end{aligned}

Because (dldt)=2vsin(θ/2)\left|\vec{\left(\dfrac{\mathrm{d}l}{\mathrm{d}t}\right)}\right|=2v\sin (\theta/2) and (dldt)×a=d \left|\vec{\left(\dfrac{\mathrm{d}l}{\mathrm{d}t}\right)} \times \vec{a}\right| = d

Now, dQ=Q2πdθ\mathrm{d}Q=\dfrac{Q}{2\pi}\mathrm{d}\theta.

So, B=μo4π2Qvd(r2+d2)3/202πsin(θ/2)dθ=μoQvdπ2(r2+d2)3/2\begin{aligned} B &=\dfrac{\mu_o}{4\pi^2} \dfrac{Qvd}{(r^2+d^2)^{3/2}} \int_{0}^{2\pi} \sin (\theta/2) \mathrm{d}\theta\\ &=\boxed{\dfrac{\mu_o Qvd}{\pi^2 (r^2+d^2)^{3/2}}} \end{aligned}

Note by Pratik Shastri
4 years, 11 months ago

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Hi Pratik,

Can you explain where the sinθ/2\sin\theta/2 originates in your derivation? On the left hand side you seem to have a scalar BB field (one component), while on the right you have a vector expression. When I calculate the components, I have some that cancel when integrated around the circle, and a zz-component that survives. The zz-component of the cross product gives (a×dl)z=r2dθ\left(\vec{a}\times d\vec{l}\right)_z = r^2d\theta.

Josh Silverman Staff - 4 years, 11 months ago

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dldt\dfrac{\mathrm{d}l}{\mathrm{d}t} is the velocity of the infinitesimally small element dθ\mathrm{d}\theta. The distance of this element from the instantaneous axis of rotation can be found using simple geometry. It comes out to be 2rsin(θ/2)2r\sin (\theta/2). Also, dl×a\vec{\mathrm{d}l}\times \vec{a} is simply dd as dl\vec{\mathrm{dl}} is always in the plane of the ring.

Pratik Shastri - 4 years, 11 months ago

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Yes, I found my mistake, I didn't check the xyx-y components.. but still why will do they cancel out? The top part of the wheel moves faster than the bottom part, right?

Pratik Shastri - 4 years, 11 months ago

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Are you thinking in the reference frame of the wheel or the frame of the ground?

Josh Silverman Staff - 4 years, 11 months ago

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@Josh Silverman The ground.

Pratik Shastri - 4 years, 11 months ago

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