91Ó°ÊÓ

x-dimension${\text{d}}_{\text{x}}=\text{5.00″¾}$

y-dimension${\text{d}}_{\text{y}}=\text{3.00″¾}$

z-dimension

The velocity of solid is

Magnetic field is

Use the concept of Lorentz force and the concept of potential across the plates. The velocity of the solid is constant, which means the forces are balanced. Using the equations, find the electric field and the potential difference.

Formulae are as follows:

$\stackrel{→}{F}=q\stackrel{→}{E}+q(\stackrel{→}{v}×\stackrel{→}{B})$

$Ed=V$

Where F 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.

The electric field within the solid in unit vector notation:

As the solid is in uniform motion, the forces are balanced, so the net force will be zero.

It can be written as,

$\begin{array}{l}0=q\stackrel{→}{E}+q(\stackrel{→}{v}×\stackrel{→}{B})\\ q\stackrel{→}{E}=−q(\stackrel{→}{v}×\stackrel{→}{B})\\ \stackrel{→}{E}=−(\stackrel{→}{v}×\stackrel{→}{B})\\ \stackrel{→}{E}=(20.0\text{ }m/s)×(30.0×{10}^{−3}\text{ }T)(−\widehat{i}×\widehat{j})\\ \stackrel{→}{E}=−(0.600\text{ }V/m)\widehat{k}\end{array}$

Hence, the electric field within the solid in unit vector notation is, $\stackrel{→}{E}=−(0.600\text{ }V/m)\widehat{k}$.

The potential difference across the solid:

The electric field is along,${\text{d}}_{\text{z}}$ so the potential difference is,

$\begin{array}{c}\text{V}=E{d}_z\\ =(0.600\text{ }V/m)×2.00\text{ }m\\ =1.20\text{ }V\end{array}$

Hence, the potential difference across the solid is,.$\text{V}=\text{1.20V}$

Therefore, use the equation of Lorentz force and the concept of potential across the plates to find the electric field and potential difference.