91Ó°ÊÓ

$\text{V}=3.9\text{‰ӼV}\left(\frac{{10}^{−6}\text{ V}}{1\text{‰ӼV}}\right)=3.\text{9}×{\text{10}}^{\text{-6}}\text{ V}$

$\text{d}=\text{0.850 c³¾}\left(\frac{{10}^{−2}\text{″¾}}{1\text{ c³¾}}\right)=8.50×{\text{10}}^{\text{-3}}\text{m}$

$\text{B}=1.20\text{ }\mathrm{mT}\left(\frac{{\text{10}}^{\text{-3}}\text{T}}{1\text{ }\mathrm{mT}}\right)=1.\text{20}×{\text{10}}^{\text{-3}}\text{T}$

If the strip is moving with constant velocity, then acceleration will be zero. So electric and magnetic forces will balance each other

Formulae are as follows:

$F_e=qE$

$F_m=qvB$

$\text{E}=\text{V/d}$

Where F_eis electric force, F_m is a magnetic force,

v is velocity, E is the electric field, B is the magnetic field, q is the charge of the particle, and d is distance.

Here, both forces are in balance.

Hence,

${\text{F}}_{\text{m}}={\text{F}}_{\text{e}}$

$\text{qvB}=\text{qE}$

$\text{v}=\frac{\text{E}}{\text{B}}$

$\text{v}=\frac{\text{V}}{\text{Bd}}$

$\begin{array}{c}\text{v}=\frac{3.9×{10}^{−6}\text{ }V}{1.20×{10}^{−3}\text{ }T×8.50×{10}^{−3}\text{ }m}\\ =0.382\text{ }m/s\end{array}$

Hence, the speed of the metal strip is,$\text{v}=\text{0.382m/s}$.

Therefore, by using the formula of electric and magnetic forces, the velocity of the metal strip can be determined.