JEE Main202120 Jul 2021Morning ShiftPhysicsCurrent ElectricityActual
A current of 5 A is passing through a non-linear magnesium wire of cross-section 0 . 04 m 2 . 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 ρ = 44 × 10 - 8 Ωm )
Options
- A11 × 10 - 2   V   m - 1
- B11 × 10 - 7   V   m - 1
- C11 × 10 - 5   V   m - 1
- D11 × 10 - 3   V   m - 1
Correct answer
C. 11 × 10 - 5   V   m - 1
Step-by-step solution
I = J → · A → = J A cos ( θ ) 5 = J 4 100 × cos ( 60 ) J = 5 × 50 = 250   A   m - 2 Now, E → = ρ × J → = 44 × 10 - 8 × 250 = 11 × 10 - 5   V   m - 1