Question

A rigid circular loop of radius r and mass m lies in the x-y plane on a flat table and has a current I flowing in it. At this particular place. The Earth’s magnetic field is $\overrightarrow B = Bxi + Byj$. The minimum value of I for which one end of the loop will lift from the table is:
A) $\dfrac{{mg}}{{\pi rBx}}$ ;
B) $\dfrac{{mg}}{{\pi x\beta y}}$;
C) $\dfrac{{mg}}{{2\pi r\sqrt {B_x^2 + B_y^2} }}$
D) $\dfrac{{mg}}{{\pi r}}\sqrt {B_x^2 + B_y^2} $

Answer 1

Hint: Here, we need to use magnetic torque which is used to put a force on the current carrying wire loop, this magnetic torque must oppose the gravitational torque so that the current carrying wire lifts its one end.

Complete step by step solution:
To lift the one end of the loop the magnetic torque must be equal to the gravitational torque:
${\tau _{magnetic}} = {\tau _{gravitational}}$;
Here the magnetic torque is given as:
${\tau _{magnetic}} = I\pi {R^2}B;$
Here:
B = Magnetic Field;
I = Current;
R = Radius;
${\tau _{gravitational}} = mgR$;
Here:
m = Mass;
g = gravitational acceleration;
R = Radius;
Equate the above two together:
$I\pi {R^2}B = mgR$;
Solve for I:
\[ \Rightarrow I = \dfrac{{mgR}}{{\pi {R^2}B}}\];
Cancel out the common factors:
\[ \Rightarrow I = \dfrac{{mg}}{{\pi RBx}}\];

Option (A) is correct, Therefore, the minimum value of I for which one end of the loop will lift from the table is $\dfrac{{mg}}{{\pi rBx}}$.

Note: Here, the loop is in two dimensions, before the loop rises it is in x direction, so the minimum amount of current that is required to lift just one end of the loop would be in the x direction only and not in the y direction. Here one end of the loop is being lifted the magnetic torque or field must be in the opposite direction to make the loop lift.