A wooden disc of mass $M = π k g$ and radius R = 50 cm is mounted on a massless rod of length d = 40 cm. The disc has a single loop of current carrying wire wound on its circumference. The ends of the rod are supported at its end so that rod is horizontal and disc is vertical. A uniform magnetic field B = 1 tesla exist in vertically upward direction. Find the least value of current (ampere) in the wire for which one end of the rod leaves the support? (Take $g = 10 m / s^{2}$ )

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Detailed Solution

$B π R^{2} I \geq M g \frac{d}{2}$

$I \geq \frac{M g d}{2 π R^{2} B}$

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