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 38-E21). 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?

Answer 1

Given: Blv =mg

When wire is replace, we have

2mg -Blv =2ma..........(i)

$\Rightarrow a=\frac{2mg-Blv}{2m}$

Now, $s=ut+\frac{1}{2}a{t}^2$

$\Rightarrow l=\frac{1}{2}\times \frac{2mg-Blv}{2m}\times t^2$

$[\because s=l]$

$t=\sqrt{\frac{4ml}{2mg-Blv}}$

$=\sqrt{\frac{4ml}{2mg-mg}}$ [From(1)]

$=\sqrt{\frac{2l}{g}}$