A square current carrying loop made of thin wire and having a mass `m=10g` can rotate without friction with respect to the vertical axis `OO_(1)` passing throught the centre of the loop at right angles to two opposite sides of the loop. The loop is placed in a homogeneous magnetic field with and induction `B =10^(-1)T` directed at right angles to the plane of the drawing `A` current `I=2A` is flowing in the loop Find the period of small oscillations that the loop performs about its position of stable equilibrium

Answer

`T_(0) =2 pi sqrt((m)/(6IB)) =0.57 s`

Explanation

`tau = MB sin theta = Ia^(2) Bsintheta= Ia^(2) Btheta`
comparing with `tau = -Ctheta`
`T =2pi sqrt((I)/(C)) since I =(ma^(2))/(6) so T = 2pi sqrt((ma^(2))/(6 xx Ia^(2)B))`
`=2pi sqrt((m)/(6IB))`
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