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Questions  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Magnetism and Matter
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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

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

 

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