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# 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

I2

\begin{align} \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{align}

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

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

So, \begin{align} 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{align}

Note by Pratik Shastri
2 years, 5 months ago

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Hi Pratik,

Can you explain where the $$\sin\theta/2$$ originates in your derivation? On the left hand side you seem to have a scalar $$B$$ 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 $$z$$-component that survives. The $$z$$-component of the cross product gives $$\left(\vec{a}\times d\vec{l}\right)_z = r^2d\theta$$. Staff · 2 years, 5 months ago

Yes, I found my mistake, I didn't check the $$x-y$$ components.. but still why will do they cancel out? The top part of the wheel moves faster than the bottom part, right? · 2 years, 5 months ago

Are you thinking in the reference frame of the wheel or the frame of the ground? Staff · 2 years, 5 months ago

The ground. · 2 years, 5 months ago

$$\dfrac{\mathrm{d}l}{\mathrm{d}t}$$ is the velocity of the infinitesimally small element $$\mathrm{d}\theta$$. The distance of this element from the instantaneous axis of rotation can be found using simple geometry. It comes out to be $$2r\sin (\theta/2)$$. Also, $$\vec{\mathrm{d}l}\times \vec{a}$$ is simply $$d$$ as $$\vec{\mathrm{dl}}$$ is always in the plane of the ring. · 2 years, 5 months ago