A rigid circular loop of radius 3.5m and mass 11kg lies in x−y plane on a smooth horizontal plane. A current flow through it. The earth’s magnetic field is B→=5i^+8k^(T) at this place. Find the value of current so that one edge of the loop is lifted from the table. g=10ms-2

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A rigid circular loop of radius $3.5 m$ and mass $11 k g$ lies in $x - y$ plane on a smooth horizontal plane. A current flow through it. The earth’s magnetic field is $\vec{B} = 5 \hat{i} + 8 \hat{k} (T)$ at this place. Find the value of current so that one edge of the loop is lifted from the table. $(g = 10 {ms}^{- 2})$

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answer is 2.

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Detailed Solution

In critical condition, torque on loop is equal to gravitational torque about an axis tangent to the loop, $τ = m g r$ , $B_{x}$ is contributing a torque,

$\begin{array}{l} τ = | \vec{M} \times \vec{B} | = M B \sin 90^{0} = π r^{2} i B_{x} \\ i = \frac{m g}{π r B_{x}} = (\frac{7}{22}) (\frac{11 \times 10}{3.5 \times 5}) = 2 A \end{array}$