Question

A cylindrical wire of radius \[0.5{\text{ mm}}\]and conductivity \[5 \times {10^7}{\text{ S/m}}\] is subjected to an electric field of \[10{\text{ mV/m}}\]. The expected value of current in the wire will be \[{x^3}\pi {\text{ mA}}\]. Find the value of \[x\].

Answer 1

Hint: In this question, we need to find the value of x. For this, we will use the formulae of current density and exponential property to get the desired result.

Formula used: The following formulae are used to solve the given question.
The current density of a conductor \[\left( J \right)\] is given by
\[J = \sigma \times E\]
Here, \[\sigma \] is the conductivity
\[E\] is the electric field
Also, the current flowing through the conductor is given by
\[I = J \times \pi {r^2}\]
Here, \[J\] is the current density of a conductor
\[r\] is the radius
Here, we will use the exponential property such as \[{a^m} \times {a^n} = {a^{m + n}}\]

Complete step by step solution:
We know that the radius \[\left( r \right)\] of cylindrical wire is 0.5 mm.
But \[1{\text{ m }} = {10^3}{\text{ mm}}\]
Thus, we get
\[r = \dfrac{{0.5}}{{{{10}^3}}} \Rightarrow 0.5 \times {10^{ - 3}}{\text{ m}}\]
That is \[r = 5 \times {10^{ - 4}}{\text{ m}}\]
Now, electric field is \[10{\text{ mV/m}}\]
Thus, we get \[10{\text{ mV/m}} = 10 \times {10^{ - 3}}{\text{ V/m}}\]
Also, the conductivity \[\left( \sigma \right)\] is \[5 \times {10^7}{\text{ S/m}}\]

Let us find the current density of a conductor.
So, we get
\[J = \sigma \times E\]
Put \[\sigma = 5 \times {10^7}{\text{ S/m}}\] and \[E = 10 \times {10^{ - 3}}{\text{ V/m}}\] in the above equation.
So, we get
\[J = 5 \times {10^7} \times 10 \times {10^{ - 3}}\]
By simplifying, we get
\[J = 5 \times {10^{7 - 3 + 1}}\]
\[\Rightarrow J = 5 \times {10^5}{\text{ A/}}{{\text{m}}^2}\]
Now, we will find the current flowing through the conducting wire.
\[I = J \times \pi {r^2}\]
Put \[J = 5 \times {10^5}{\text{ A/}}{{\text{m}}^2}\] and \[r = 5 \times {10^{ - 4}}{\text{ m}}\] in the above equation.

So, we get
\[I = 5 \times {10^5} \times \pi {\left( {5 \times {{10}^{ - 4}}} \right)^2}\]
\[\Rightarrow I = 5 \times {10^5} \times \pi \left( {25 \times {{10}^{ - 8}}} \right)\]
By simplifying, we get
\[I = 125 \times {10^{5 - 8}} \times \pi \]
\[\Rightarrow I = 125 \times {10^{ - 3}} \times \pi \]
That is \[I = 125 \times \pi {\text{ mA}}\]
By comparing the above equation with \[{x^3}\pi {\text{ mA}}\], we get
\[{x^3} = 125\]
By taking the cube root on both sides, we get
\[x = 5\]

Therefore, the value of x is 5.

Note: Many students generally make mistakes in writing the formula of current density and current flows through the conductor. Also, they may get wrong while converting units such as mm to m etc. If so, the end result may get wrong.