Figure 28-35 shows a metallic block, with its faces parallel to coordinate axes. The block is in auniform magnetic field of magnitude 0.020 T. One edge length of the block is 25 cm; the block is not drawn to scale. The block is moved at 3.0 m/s parallel to each axis, in turn, and the resulting potential difference Vthat appears across the block is measured. With the motion parallel to the y-axis, V= 12 mV; with the motion parallel to the z-axis, V= 18 mV; with the motion parallel to the x-axis, V= 0. What are the block lengths (a) d_x, (b) d_y, and (c) d_z?

The magnetic field is.$\text{B}=\text{0.020T}$

One edge length of the block is $\text{L}=\text{25cm}$.

Speed of block along x is .${\text{v}}_{\text{x}}=\text{3.0m/s}$

Voltage along y is ${\text{V}}_{\text{y}}=\text{12mV}\left(\frac{{10}^{-3}V}{1mV}\right)=0.012V$.

Voltage along z is ${\text{V}}_{\text{z}}=\text{18mV}\left(\frac{{10}^{-3}V}{1mV}\right)=0.018V$

Voltage along x is ${\text{V}}_{\text{x}}=\text{0mV}$

Use the concept of voltage across two plates. Using the equation of voltage related to electric field and distance, find the distance.

Formulae are as follows:

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

Where E is the electric field, V is potential, and d is the distance between plates.

Length of block${\text{d}}_{\text{x}}$:

The distance along x is given${\text{d}}_{\text{x}}=\text{0.25m}\left(\frac{100cm}{1m}\right)$ .

Hence, the length of the block ${\text{d}}_{\text{x}}=\text{25cm}$.

Write the equation for electric field from electric force and magnetic force.

$qE=qvB$

$E=vB$

Using this value in the equation of voltage,

$vB=\frac{V}{d}$

Rearranging for d,

$d=\frac{V}{vB}$

It is given that v, d, and B are mutually perpendicular to each other.

Whenvelocity is along y, then the voltage is$\text{0.012V.}$

$d_y=\frac{V_y}{v_{y}B}$

$\begin{array}{l}{d}_y=\frac{0.012V}{3.0m/s\times 0.020T}\\ d_y=0.20m\left(\frac{100cm}{1m}\right)\\ d_y=20cm\end{array}$

Hence, the length of the block .${\text{d}}_{\text{y}}=2\text{0cm}$

Length of block${\text{d}}_{\text{z}}$:

${\text{d}}_{\text{z}}=\frac{{\text{V}}_{\text{z}}}{{\text{v}}_{\text{z}}\text{B}}$

$\begin{array}{l}{\text{d}}_z=\frac{0.018V}{3.0m/s\times 0.020T}\\ {\text{d}}_z=0.30m\left(\frac{100cm}{1m}\right)\\ {\text{d}}_z=30cm\end{array}$

Hence, the length of the block ${\text{d}}_{\text{z}}=3\text{0cm}$

Therefore, use the concept of voltage across two plates to find the distances.