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  • A circular loop of mass 10kg and radius 5m is kept in a horizontal position (XY plane) in a table. A uniform magnetic field B is applied parallel to x-axis. The current in the loop so that its one

A circular loop of mass 10kg and radius 5m is kept in a horizontal position (XY plane) in a table. A uniform magnetic field B is applied parallel to x-axis. The current in the loop so that its one point just lifts from the table is:

 

Option: 1

\frac{10g}{\pi B}


Option: 2

\frac{25g}{\pi B}


Option: 3

\frac{2g}{\pi B}


Option: 4

\frac{g}{\pi B}


Answers (1)

best_answer

The magnetic moment of the loop is:

m = \pi r^2 l = \pi * (5)^{2} * l = 25 \pi l

and

\left| \vec{r }\right| = \left| \vec{u} \times \vec {B}\right|

\implies r = \pi r^{2} l B

For the loop to be just lifted,

mg r = \pi r^{2} l B

\implies I = \frac{mg}{\pi r B} = \frac{10g}{5 \pi B} = \frac{2g}{\pi B}

Posted by

rishi.raj

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