A) \[\,\frac{mg}{\pi r\sqrt{B_{x}^{2}+B_{z}^{2}}}\]
B) \[\frac{mg}{\pi r{{B}_{z}}}\]
C) \[\frac{mg}{\pi r{{B}_{x}}}\]
D) \[\frac{mg}{\pi r\sqrt{{{B}_{x}}{{B}_{z}}}}\]
Correct Answer: C
Solution :
The torque on the loop must be equal to the gravitational torque exerted about an axis tangent to the loop. |
The gravitational torque: |
\[{{\tau }_{1}}=mgr\] ...(1) |
Only \[{{B}_{x}}\], causes a torque. Therefore torque to the magnetic field |
\[{{\tau }_{2}}=\left| \overrightarrow{M}\times \overrightarrow{B} \right|=MB\sin 90{}^\circ =\pi {{r}^{2}}i{{B}_{x}}\] ...(2) |
Eqs.(1) and (2),we get, \[i=\frac{mg}{\pi r{{B}_{x}}}\] |
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