Question: In Fig. 28-57, the two ends of a U-shaped wire of mass m=10.0gand length L=20.0cmare immersed in mercury (which is a conductor).The wire is in a uniform field of magnitude B=0.100T. A switch (unshown) is rapidly closed and then reopened, sending a pulse of current through the wire, which causes the wire to jump upward. If jump height h=3.00m, how much charge was in the pulse? Assume that the duration of the pulse is much less than the time of flight. Consider the definition of impulse (Eq. 9-30) and its relationship with momentum (Eq. 9-31). Also consider the relationship between charge and current (Eq. 26-2).

$m=10\text{g}=0.01\text{kg}$

$L=20cm=0.2m$

$$\begin{gathered} B=0.1\text{T} \\ h=3\text{m} \end{gathered}$$

By using the magnetic force on a shaped wire, first we consider momentum, which is mass times velocity. From that, we calculate the velocity of a charged particle. By using the law of conservation of energy, the total change in kinetic energy is equal to the change in potential energy. So by equating this equation, we can find the charge in the pulse.

Formula:

$F=iBL$

$\Delta p=m\Delta v$

As we know, the force due to U shaped wire is

$F=BiL$

Where B is the magnetic field, L is the length of the wire, and current is i.

Now we know momentum as

$$\begin{gathered} \Delta p=m\Delta v \\ \Delta p=\int Fdt \\ \Delta p=\int BiLdt \\ \Delta p=BL\int idt \end{gathered}$$

But the$\int idt$is the total charge q

So that$\Delta p=BLq$

Putting the value of momentum as

$$\begin{gathered} mv=BLq \\ v=\frac{BLq}{m}······\left(1\right) \end{gathered}$$

But the change in kinetic energy is the change in potential energy as

$$\begin{gathered} \frac{1}{2}m{v}^2=mgh \\ \frac{1}{2}{v}^2=gh \end{gathered}$$

Putting the value of v as from equation (1), we can write

$\frac{1}{2}{\left(\frac{BLq}{m}\right)}^2=gh$

From that, we can find the relation for charge as

$q=\sqrt{2gh}\frac{m}{BL}$

By substituting the value, we can find the charge in pulse as

$$\begin{gathered} q=\sqrt{2\times 9.8\times 3}\times \frac{0.01}{0.1\times 0.2} \\ q=3.83\text{C} \end{gathered}$$

Hence, the charge in the pulse is$q=3.83\text{C}$