$\text{PQ}=\text{RS}=1$
$\text{PS}=\text{QR}=\text{b}$
Area $\text{A}=\text{lb}$
$\vec{\text{M}}\times\vec{\text{IA}}$
$\overrightarrow{\text{F}_{\text{PQ}}}=\text{IlB}\otimes$
$\overrightarrow{\text{F}_{\text{RS}}}=\text{IlB}\ominus$
$\overrightarrow{\text{F}_{\text{QR}}}=\text{IbB}\sin(90^\circ-\theta)=\text{IbB}\cos\theta\text{ up}$
$\overrightarrow{\text{F}_{\text{SP}}}=\text{IbB}\sin(90^\circ-\theta)=\text{IbB}\cos\theta\text{ down}$
Only $\overrightarrow{\text{F}_{\text{PB}}}\ \& \ \overrightarrow{\text{F}_{\text{RS}}}$ form a couple to apply torque on loop,
$\tau=\text{MB}\sin\theta$
Magnetic field is taken radial in Galvanometer coil in order to create $\theta=90^\circ$ at every orientation of coil in the magnetic field so that current varies linearly with deflection.
$\Rightarrow\text{v}=\sqrt{\frac{2\text{qV}}{\text{m}}}$
$\because\vec{\text{v}}=\text{vi}\perp\vec{\text{B}}(=\text{Bj})$
$\therefore$ Particle deflects along circular path of radius $\text{r}=\frac{\text{mv}}{\text{qB}}=\frac{\text{m}}{\text{qB}}\sqrt{\frac{2\text{qv}}{\text{m}}}=\frac{1}{\text{B}}\sqrt{\frac{2\text{mv}}{\text{q}}}$
$\text{r}=\frac{1}{2\times10^{-3}}\sqrt{\frac{2\times6.4\times10^{-27}\times10^4}{2\times1.6\times10^{-19}}}$
$=\frac{1}{2\times10^{-3}}\times2\times10^{-2}$
$=10^1\text{m}=10\text{m}.$
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
