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?

Answer 1

Let us consider the above free body diagram.
As the net force on the wire is zero, ilB = mg.
When the wire is replaced by a wire of double mass, we have



Now, let a' be the acceleration of the wire in downward direction and t be the time taken by the wire to fall.
Net force on the wire = 2mg − ilB = F_net
On applying Newton's second law, we get
2mg − ilB = 2 ma' ...(1)
$$\begin{gathered} \Rightarrow a\text{'}=\frac{2mg-ilB}{2m} \\ s=ut+\frac{1}{2}a\text{'}{t}^2 \\ \Rightarrow l=\frac{1}{2}\times \frac{2mg-ilB}{2m}\times t^2[\because s=l] \\ \Rightarrow t=\sqrt{\frac{4ml}{2mg-ilB}} \\ \Rightarrow t=\sqrt{\frac{4ml}{2mg-mg}}[\mathrm{From}(1\left)\right] \\ \Rightarrow t=\sqrt{\frac{2l}{g}} \end{gathered}$$