A uniform rigid ring of radius r and unknown mass is placed on a non-conducting frictionless horizontal tabletop, where exists a uniform horizontal magnetic field of induction B as shown in the figure. A current I that is double of the minimum current required to flip over the ring is made to circulate in the coil. Find force (in N) of normal reaction from the tabletop on the ring immediately after the current is switched on. Take, $r = \frac{1}{2} m; B = \frac{6}{π} T; I = 2 A$

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Detailed Solution

${\bar{τ}}_{m} = \bar{μ} \times \bar{B}$ $τ_{m} = i . π R^{3} . B$

Minimum current to flip the ring:

$i π R^{2} B \geq m g R \Rightarrow i_{\min} = \frac{m g}{π R B}$ If current to flip the ring:

$I = 2 i_{\min} \Rightarrow τ_{m} = 2 m g R ∴ α = \frac{2 m g - m g}{\frac{3}{2} m R^{2}} = \frac{2 g}{3 R} \Rightarrow a_{c} = \frac{2 g}{3} ∴ N - m g = m a_{c} \Rightarrow N = m g + \frac{2 m g}{3} \Rightarrow N = \frac{5}{3} m g = \frac{5}{3} π R B i_{\min} = \frac{5}{6} π R B I$