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A conducting ring of radius R & mass 300 g, carrying current I ampere as shown in the figure. A bar magnet with its north pole up is placed along the symmetric axis below the conducting ring at a distance of 2m from the centre of ring. At the location of ring, the magnetic field makes an angle $30^{0}$ with vertical and its magnitude is $\frac{1}{3 π}$ Tesla. If ring remains at rest in its position, find the product of current I in the ring and radius R of ring in ampere-meter $(g = 10 m / s^{2}) .$

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answer is 9.

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Detailed Solution

$F_{m a g} = m g \Rightarrow (B \sin 30^{0}) \times I \times (2 π R) = m g \Rightarrow I R = \frac{m g}{2 π B \sin 30^{0}} = \frac{0.3 \times 10 \times 3 π \times 2}{2 π} = 9 A - m$

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