Question

A current of $5\text{A}$ is passing through a non-linear magnesium wire of cross-section $0.04{\text{m}}^2$. At every point, the direction of current density is at an angle of ${60}^{\circ }$, with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is :

Resistivity of magnesium is $44\times {10}^{–8}\Omega \text{-m}$.

• A. $11\times {10}^{-7}\text{V/m}$
• B. $11\times {10}^{-3}\text{V/m}$
• C. $11\times {10}^{-5}\text{V/m}$
• D. $11\times {10}^{-2}\text{V/m}$

Answer 1

The correct option is C $11\times {10}^{-5}\text{V/m}$

We know that,

$I=\overrightarrow{J}.\overrightarrow{A}=JA\cos \theta$

$\Rightarrow 5=J\times 0.04\times \cos {60}^{\circ }$

$\Rightarrow J=250{\text{A/m}}^2$

Now,

$J=\sigma E$

$\Rightarrow E=\rho J$

$\Rightarrow E=44\times {10}^{-8}\times 250=11\times {10}^{-5}\text{V/m}$

Hence, option $\left(C\right)$ is the correct answer.