JEE Main 2021 — Current Electricity Question with Solution
From: JEE Main 2021 (Online) 20th July Morning Shift
Question
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 with the unit vector of area of cross-section. The magnitude of electric field at every point of the conductor is :
(Resistivity of magnesium = 44 108 m)
(Resistivity of magnesium = 44 108 m)
Choose an option
Show full solutionCorrect option: A
Correct answer
A11 105 V/m
Step-by-step explanation
Given, current, I = 5A
Area of cross-section of wire, A = 0.04 m2
We know that,
or or
where, J = current density.
[ Given, = 60]
[ cos60 = ]
J = 250 Am2
The relation between electric field, current density and resistivity can be given as,
E = . J
= 44 108 250 [ Resistivity, = 44 108 -m]
= 11 105 V/m
Area of cross-section of wire, A = 0.04 m2
We know that,
or or
where, J = current density.
[ Given, = 60]
[ cos60 = ]
J = 250 Am2
The relation between electric field, current density and resistivity can be given as,
E = . J
= 44 108 250 [ Resistivity, = 44 108 -m]
= 11 105 V/m
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