Question

In a circuit there is a copper wire 40 cm long with a potential difference from one end to the other end of $\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{V}$. What is the magnitude of electric field inside the wire?

Short answer

$0.025V/m$

Step-by-step solution

Identification of the given data

The given data is listed below,

• The length of the copper wire is, $\mathrm{L}=40\mathrm{cm}\times \frac{1\mathrm{m}}{100\mathrm{cm}}=0.4\mathrm{m}$
• The potential difference across the wire is, $∆V=0.01v$

Significance of the electric field

The electric field is a region in which a charged particle is able to exert force on another charged particle.

The ratio of potential difference across an object to the length of the object gives the value of the electric field.

Determination of the magnitude of the electric field

The equation of the magnitude of the electric field is expressed as,

$E=\frac{∆V}{L}$

Here,$∆V$is the potential difference across the wire and $L$ is the length of the wire.

Substitute all the values in above expression.

$$\begin{gathered} \mathrm{E}=\frac{0.01\mathrm{V}}{0.4\mathrm{m}} \\ =0.025\mathrm{V}/\mathrm{m} \end{gathered}$$

Thus, the magnitude of the electric field is $0.025\mathrm{V}/\mathrm{m}$.