Question

A metal strip6.50cm long, 0.850 cm wide, and 0.760 mm thick moves with constant velocity through a uniform magnetic field B=1.20mT directed perpendicular to the strip, as shown in Fig.12-34. A potential difference of 3.90$\mu v$is measured between points xand yacross the strip. Calculate the speed.

Short answer

The speed of the metal strip is$v=3.9\mu V\left(\frac{{10}^{-8}}{1\mu v}\right)=3.9\times {10}^{-6}v$ v=0.382 m/s .

Step-by-step solution

Given

$\text{V}=3.9\text{\hspace{0.17em}μV}\left(\frac{{10}^{-6}\text{\hspace{0.17em}V}}{1\text{\hspace{0.17em}μV}}\right)=3.\text{9}\times {\text{10}}^{\text{-6}}\text{\hspace{0.17em}V}$

$\text{d}=\text{0.850\hspace{0.17em}cm}\left(\frac{{10}^{-2}\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}cm}}\right)=8.50\times {\text{10}}^{\text{-3}}\text{m}$

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

Determining the concept

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.

Determining the speed of the metal strip 

Here, both forces are in balance.

Hence,

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

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

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

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

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

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.