A current of 5 A is passing through a non-linear magnesium wire of cross-section ${\rm{0}}{\rm{.04 }}\,\,{{\rm{m}}^{\rm{2}}}$ At every point, the direction of current density is at an angle of ${60^{\rm{o}}}$ with the unit vector of area of cross -section. The magnitude of electric field at every point of the conductor is (Take resistivity of magnesium, ${\rm{\rho }} = 44 \times {10^{ - 8}}\,\Omega - m)$

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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A current of 5 A is passing through a non-linear magnesium wire of cross-section

\mathrm{0}\mathrm{.04 }\,\,{\mathrm{m}^\mathrm{2}}

At every point, the direction of current density is at an angle of

{60^\mathrm{o}}

with the unit vector of area of cross -section. The magnitude of electric field at every point of the conductor is
(Take resistivity of magnesium,

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

View answer and explanation

Correct option: C

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

Option C (

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

) is correct because it is calculated using the relationship between electric field, current density, and resistivity, considering the angle of the current density.

JEE PHYSICS MCQ: CURRENT ELECTRICITY

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