Question

A circular loop of mass $m$ and radius $r$ lies on a horizontal table(xy-plane). A uniform magnetic field is applied parallel to the x-axis the current $I$ that should flow in the loop so that it just tilts about one point on the table is :

• A. $\frac{mg}{{\pi }^{2}B}$
• B. $\frac{mg}{2\pi rB}$
• C. $\frac{mg}{\pi rB}$
• D. $\frac{\pi rB}{mg}$

Answer 1

Answer: (C)

$\frac{mg}{\pi rB}$

If the loop just tilts about one point on the table, the forces acting on it is mg downward where m is the mass of the loop and g is acceleration due to gravity. mg will provide a torque about the point of contact with the table.
Torque due to weight $(\tau {)}_{mg}=\overrightarrow{r}\times (m\overrightarrow{g})$
The circular current carrying loop can be treated like a magnet and will have a magnetic moment $\overrightarrow{\mu }=i\overrightarrow{A}$
Torque on the loop due to magnetic field $\overrightarrow{B}$ is: $\overrightarrow{\mu }\times \overrightarrow{B}=i\overrightarrow{A}\times \overrightarrow{B}=i\left(\pi r^2\right)\hat{A}\times \overrightarrow{B}$
Both the torques will balance.
$\therefore i\left(\pi r^2\right)\hat{A}\times \overrightarrow{B}=\overrightarrow{r}\times \left(m\overrightarrow{g}\right)$
$\therefore i\left(\pi r^2\right)B=mgr$
$\Rightarrow i=\frac{mgr}{\pi r^{2}B}=\frac{mg}{\pi rB}$