Question

A current of 1.5 A is maintained in a copper wire of length 1 m with area of cross-section $1.7 \times 10^{-7}$ m$^2$. The magnitude of electric field in the wire is $[\rho_{Cu} = 1.7 \times 10^{-8}\,\Omega\,\text{m}]$:

• A. 0.15 V/m
• B. 0.30 V/m
• C. 1.5 V/m
• D. 3.0 V/m

Hint:

Use \(E = \rho I / A\). The numbers \(1.7 \times 10^{-7}\) cancel, leaving \(E = 1.5\) V/m.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Inside a conductor, the current density is \(J = \dfrac{I}{A}\), and the electric field is related to it by \(E = \rho J\). This is the microscopic form of Ohm's law.

Step 2: Key Formula:
\[ E = \rho \frac{I}{A} \]
Here \(\rho = 1.7 \times 10^{-8}\ \Omega\,\text{m}\), \(I = 1.5\) A and \(A = 1.7 \times 10^{-7}\ \text{m}^2\).

Step 3: Substitute the Values:
\[ E = \frac{(1.7 \times 10^{-8})(1.5)}{1.7 \times 10^{-7}} \]
The numbers \(1.7\) cancel, and \(10^{-8}/10^{-7} = 0.1\). So \(E = 0.1 \times 1.5 = 0.15\) V/m.
Units check: \(\Omega\,\text{m} \times \text{A} / \text{m}^2 = \text{V/m}\).

Step 4: Why the Other Options Are Wrong:
0.30 V/m and 3.0 V/m would need \(\rho\) to be 2 times and 20 times larger than the given value. 1.5 V/m comes from using \(\rho = 1.7 \times 10^{-7}\ \Omega\,\text{m}\) by mistake, which is ten times too large. The length of 1 m does not change the field in a uniform wire.

Final Answer:
The electric field in the wire is 0.15 V/m, option (A). \[ \boxed{0.15\ \text{V/m}} \]