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

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

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, \[\begin{align} B &=\dfrac{\mu_o}{4\pi^2} \dfrac{Qvd}{(r^2+d^2)^{3/2}} \int_{0}^{2\pi} \sin (\theta/2) \mathrm{d}\theta\\ &=\boxed{\dfrac{\mu_o Qvd}{\pi^2 (r^2+d^2)^{3/2}}} \end{align}\]

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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?

Log in to reply

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

Log in to reply

Log in to reply

\(\dfrac{\mathrm{d}l}{\mathrm{d}t}\) 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.

Log in to reply