91Ó°ÊÓ

If the value of the potential difference measured between the two ends of the wire is positive, then the value of the electric field would be negative and opposite to the direction of potential difference.

The electric field is also affected by the length of the wire, by increasing the wire length between two ends, electric field decreases.

The potential difference from one end to the other end of the wire is, $\mathrm{Δ³Õ}={\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}}=1.5 \mathrm{V}$.

The distance between two ends of the wire is, ${\mathrm{Δ\pm ô}}_{\mathrm{AB}}=1 \mathrm{cm}×\frac{1 \mathrm{m}}{100\mathrm{cm}}=0.01 \mathrm{m}$.

The potential difference between two ends of the wire is given by,

$$\begin{gathered} \mathrm{Δ³Õ}={\mathrm{V}}_{\mathrm{B}}-{\mathrm{V}}_{\mathrm{A}} \\ \mathrm{Δ³Õ}=1.5 \mathrm{V} \end{gathered}$$

Since, the value of$\mathrm{Δ³Õ}$ is positive, the potential is higher at end B.

The formula for themagnitude of the electric field inside the wireis given by,

$\mathrm{E}=\frac{-\mathrm{Δ³Õ}}{{\mathrm{Δ\pm ô}}_{\mathrm{AB}}}$

The path of the potential difference is from initial location A to final location B.

Putting the values,

$$\begin{gathered} \mathrm{E}=\frac{-1.5 \mathrm{V}}{0,01\mathrm{m}}×\frac{1 \mathrm{N}/\mathrm{C}}{1\mathrm{V}/\mathrm{m}} \\ \mathrm{E}=-150 \mathrm{N}/\mathrm{C} \end{gathered}$$

The negative sign indicates that the path of the electric field is opposite to the potential difference.

Hence, the magnitude of the electric field is $150 \mathrm{N}/\mathrm{C}$and its direction is from B to A.