A rectangular loop ABCD of dimensions l and b, carrying a steady current is placed in uniform magnetic field as shown in fig; such that normal of the plane is at angle q with the magnetic field lines.
The force FBC and FAD on arms BC and AD are equal, opposite and along the axis of the coil, so they cancel each other.
The forces FAB and FCD are also equal and opposite, but are not collinear, so they constitute a couple, and the magnitude of the torque can be given as
$\tau = \text{F}_{AB}.\frac{\text{b}}{2}\sin\theta + \text{F}_{CD}.\frac{\text{b}}{2}\sin\theta$
Since
$|\text{F}_{AB}| = |\text{F}_{CD}| =\text{BI}\ell$
$\tau = \text{BI}\ell\times\text{b}\times\sin\theta$
$ = \text{BI}(\ell\text{b})\sin\theta$
= BI A sin $\theta$
$[\text{A} = \ell \text{b} = \text{ area of the rectangle}]$
Since magnetic moment m = I |A|
$\tau = \text{mB} \sin \theta$
In vector from $\overrightarrow{\tau} = \overrightarrow{\text{m}}\times\overrightarrow{\text{B}}$
$\text{q}v\text{B} = \frac{\text{m}v^{2}}{\text{r}}$
$\Rightarrow\text{r} = \frac{\text{m}v}{\text{qB}} = \frac{\text{p}}{\text{qB}}$
Since momentum are equal, and they have equal charges.
So, rp : rd = 1:1.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\text{mg}$
$\frac{\text{mg}}{\cos\theta}$
$\text{mg}\cos\theta$
$\text{mg}\tan\theta$
