Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class12  >>  Magnetism and Matter
0 votes

A rigid circular loop of radius $ r$ lies in the $x-y$ plane on a flat table and a current $i$ flows through it. At this place, the earth’s magnetic field is $B_x\hat i + B_z \hat k$ . The value of $i$ so that one end of the loop lifts up from the table is

$\begin {array} {1 1} (a)\;\large\frac{mg}{\pi r \sqrt{Bx^2+Bz^2}} & \quad (b)\;\large\frac{mg}{\pi r \: Bx} \\ (c)\;\large\frac{mg}{\pi r \: Bz} & \quad (d)\;\large\frac{mg}{\pi r \sqrt{Bx.BZ}} \end {array}$


Can you answer this question?

1 Answer

0 votes
The torque on the loop must be equal to the gravitational torque exerted about an axis tangent to the loop.
$|\overrightarrow M \times \overrightarrow B| = mgr$
Ans : (b)
answered Feb 23, 2014 by thanvigandhi_1

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App