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

$\begin{aligned} \mathrm{d}B &=\dfrac{\mu_o}{4\pi} \dfrac{\mathrm{d}q}{\mathrm{d}t} \dfrac{\vec{\mathrm{d}l} \times \vec{a}}{|\vec{a}|^3}\\ &=\dfrac{\mu_o}{4\pi}\mathrm{d}q \dfrac{2vr\sin(\theta/2) d}{(r^2+d^2)^{3/2}} \end{aligned}$

Because $\left|\vec{\left(\dfrac{\mathrm{d}l}{\mathrm{d}t}\right)}\right|=2v\sin (\theta/2)$ 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, $\begin{aligned} B &=\dfrac{\mu_o}{4\pi^2} \dfrac{Qvd}{(r^2+d^2)^{3/2}} \int_{0}^{2\pi} \sin (\theta/2) \mathrm{d}\theta\\ &=\boxed{\dfrac{\mu_o Qvd}{\pi^2 (r^2+d^2)^{3/2}}} \end{aligned}$

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## Comments

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TopNewestHi 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 $\left(\vec{a}\times d\vec{l}\right)_z = r^2d\theta$.

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$\dfrac{\mathrm{d}l}{\mathrm{d}t}$ 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.

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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?

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

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