Q. A rigid circular loop of radius r and mass m lies in the $x - y$ plane of a flat table and has a current $i$ flowing in it. At this particular place the earth’s magnetic field is $\mathbf{B =}B_{x}\widehat{\mathbf{i}}\mathbf{+}B_{z}\widehat{\mathbf{k.}}$ The value of $i$ so that the loop start tilting is
A) $\frac{mg}{\pi r\sqrt{B_{x}^{2} + B_{z}^{2}}}$
B) $\frac{mg}{\pi rB_{x}}$
C) $\frac{mg}{\pi rB_{z}}$
D) $\frac{mg}{\pi r\sqrt{B_{x}B_{z}}}$
Explanation
(b) Torque($\tau$) is given by
$\ \ \ \ \ \ \tau = \left| \mathbf{M} \times \mathbf{B} \right|$
Where M is magnetic dipole moment of loop given by,
$\mathbf{\ \ \ \ M} = i\mathbf{A}$
$\mathbf{\therefore M} = (i\pi r^{2})\widehat{\mathbf{k}}$
$\ \ \ \ \ \ \tau = |\mathbf{M} \times \mathbf{B| = (}\pi r^{2}iB_{x}\mathbf{)}$
Now, torque of weight will be mgr.
$mgr = \pi r^{2}iB_{x}$
$\Rightarrow \ \ i = \frac{mg}{\pi rB_{x}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.