Question

Figure shows a horizontal wire MN of length $l$ and mass m is placed in a magnetic field B. The ends of wire are bent and dipped in two bowls containing Hg which is connected to an external circuit as shown. If key is pressed for a short time $\underline{\delta t}$ due to which a charge q flows. Find the maximum height above initial level the wire MN will jump.

Answer 1

$\frac{l^2 B^2 q^2}{2gm^2}$

Answer 2

The magnetic force on the wire is $F = IlB$. The impulse delivered by this force is $J = \int F dt = lB \int I dt = lBq$. By the impulse-momentum theorem, $mv = J$, so $v = \frac{lBq}{m}$. By conservation of energy, $\frac{1}{2}mv^2 = mgh$, so $h = \frac{v^2}{2g} = \frac{1}{2g} \left(\frac{lBq}{m}\right)^2 = \frac{l^2 B^2 q^2}{2gm^2}$.