A current of 5 A is passing through a non-linear magnesium wire of cross-section 0.04 m2. 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 \rho = 44 \times 10 - 8 \Omega m)

Current Electricity PYQ — JEE Main 2021 · 20 Jul · Shift 1 · Q47 with Solution

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Correct answer: A

For a conductor in which current density $\vec J$ makes an angle $\theta$ with the area vector $\hat n$ , the current is

I = \int \vec J \cdot d\vec A

If $J$ is uniform over the cross-section, then

I = JA\cos\theta

Given:

So,

J = \frac{I}{A\cos 60^\circ}

Since $\cos 60^\circ = \frac{1}{2}$ ,

J = \frac{5}{0.04 \times \frac{1}{2}} = \frac{5}{0.02} = 250\,\text{A/m}^2

\vec E = \rho \vec J

Hence magnitude,

E = \rho J

Given resistivity of magnesium:

\rho = 44 \times 10^{-8}\,\Omega\,\text{m}

Therefore,

E = (44 \times 10^{-8})(250)

E = 11000 \times 10^{-8}

E = 11 \times 10^{-5}\,\text{V/m}

E = 11 \times 10^{-5}\,\text{V/m}

So the correct option is A.