Understanding Torque in a Magnetic Field with Loop

[annamal]

Actually that line doesn't make sense either. The LHS is a current element but the RHS is a current times a length element.

I meant
$Idl = (I(-rsin\theta d\theta), Ircos\theta d\theta, 0)$

unsure why dl has to be in terms of r and $\theta$
if so, does $\vec dl = rd\theta (-\vec i + \vec j)$

 

[annamal]

That integral makes no sense. You show dl as the integration variable but the range seems to be for an angle. Write it all in terms of $r, \theta$ instead of l and perform that cross product.

dI = (dx, dy, 0) = $(-rsin\theta d\theta), rcos\theta d\theta, 0)$
where x = $rcos\theta$ and y = $rsin\theta$

 

[haruspex]

why dl has to be in terms of r and θ

So that you can do the integral.

if so, does $\vec dl = rd\theta (-\vec i + \vec j)$

No, it is what you wrote,

$(-rsin\theta d\theta, rcos\theta d\theta, 0)$

Now substitute that for $\vec{dl}$ in $\vec F = \int_{\theta=0}^{2\pi} I\vec{dl} \times \vec B$, and replace $\vec B$ there using $\vec B = (0, B_y, -B_z)$. Then perform the cross product, then perform the integration.

 

[annamal]

So that you can do the integral.

No, it is what you wrote,

Now substitute that for $\vec{dl}$ in $\vec F = \int_{\theta=0}^{2\pi} I\vec{dl} \times \vec B$, and replace $\vec B$ there using $\vec B = (0, B_y, -B_z)$. Then perform the cross product, then perform the integration.

Ok, so $(-Irsin\theta d\theta, Ircos\theta d\theta, 0) \times (0, B_y, -B_z) = \vec dF = (-IB_z rcos\theta d\theta, -IB_z rsin\theta d\theta, -IB_y rsin\theta d\theta)$
$\vec F = (-IB_z rsin\theta, IB_z rcos\theta, IB_y rcos\theta)$ from 0 to $2\pi$ = 0?

 

[haruspex]

Ok, so $(-Irsin\theta d\theta, Ircos\theta d\theta, 0) \times (0, B_y, -B_z) = \vec dF = (-IB_z rcos\theta d\theta, -IB_z rsin\theta d\theta, -IB_y rsin\theta d\theta)$
$\vec F = (-IB_z rsin\theta, IB_z rcos\theta, IB_y rcos\theta)$ from 0 to $2\pi$ = 0?

Ok, so you found the net force is zero. But we don't care about the net force, we care about the net torque. So find the torque on an element and then integrate.

 

[annamal]

Ok, so you found the net force is zero. But we don't care about the net force, we care about the net torque. So find the torque on an element and then integrate.

What do you mean by that
$\vec r \times \vec F = (rcos\theta, rsin\theta, 0) \times (0, 0, 0)$ = 0?

 

[haruspex]

What do you mean by that
$\vec r \times \vec F = (rcos\theta, rsin\theta, 0) \times (0, 0, 0)$ = 0?

No, $\vec r=\vec r(\theta)$, so you have to find the torque on an element of the loop and then integrate.
$\tau=\int \vec r\times\vec F.d\theta$.

 

[annamal]

No, $\vec r=\vec r(\theta)$, so you have to find the torque on an element of the loop and then integrate.
$\tau=\int \vec r\times\vec F.d\theta$.

Ok, I get it.