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?
Figure

Answer 1

EMF Equals Force

Explanation:

Given Bℓv = mg …(1)  

When wire is replaced we have  

2mg – Bℓv = 2ma [where a → acceleration]

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

Now, Using Motions II Equation,

$s = ut + \frac {1}{2}at^2$

So, l = $\frac {1}{2} \times \frac {2mg - Blv}{2m} \times t^2$ (As s = l)

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

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

= $\sqrt \frac {2l}{g}$ (from (1))