The system containing the rails and the wire of the previous prob… - Vidyadip

Question

The system containing the rails and the wire of the previous problem is kept vertically in a uniform horizontal magnetic field B that is perpendicular to the plane of the rails (figure). It is found that the wire stays in equilibrium. If the wire ab is replaced by another wire of double its mass, how long will it take in falling through a distance equal to its length?

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Answer

Given Blv = mg …(1) When wire is replaced we have 2mg - Blv = 2ma [where a → acceleration]$\Rightarrow\text{a}=\frac{2\text{mg}-\text{Blv}}{2\text{m}}$
Now, $\text{s}=\text{ut}+\frac{1}{2}\text{at}^2$$\Rightarrow\text{l}=\frac{1}{2}\times\frac{2\text{mg}-\text{Blv}}{2\text{m}}\times\text{t}^2 \ \big[\therefore \ \text{s}=\text{l}\big]$
$\Rightarrow\text{t}=\sqrt{\frac{4\text{ml}}{2\text{mg}-\text{Blv}}}=\sqrt{\frac{4\text{ml}}{2\text{mg}-\text{mg}}}=\sqrt{\frac{2\text{l}}{\text{g}}}$ [from (1)]

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