Question

The current generator I_g' shown in figure, sends a constant current i through the circuit. The wire ab has a length l and mass m and can slide on the smooth, horizontal rails connected to I_g. The entire system lies in a vertical magnetic field B. The system 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)

\[\Rightarrow a' = \frac{2mg - ilB}{2 m}\]

\[s = ut + \frac{1}{2}a' t^2 \]

\[ \Rightarrow l = \frac{1}{2} \times \frac{2mg - ilB}{2m} \times t^2 .............\left[ \because s = l\right]\]

\[ \Rightarrow t = \sqrt{\frac{4 ml}{2mg - ilB}}\]

\[ \Rightarrow t = \sqrt{\frac{4 ml}{2mg - mg}} .........\text{[From (1)]}\]

` t = 2sqrt(l/g)`