91Ó°ÊÓ

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

$L=20cm=0.2m$

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

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:

$F=iBL$

$\mathrm{Δ}p=m\mathrm{Δ}v$

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

$$\begin{gathered} \mathrm{Δ}p=m\mathrm{Δ}v \\ \mathrm{Δ}p=∫Fdt \\ \mathrm{Δ}p=∫BiLdt \\ \mathrm{Δ}p=BL∫idt \end{gathered}$$

But the$∫idt$is the total charge q

So that$\mathrm{Δ}p=BLq$

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

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

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