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 6 0 ∘ 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 × 1 0 − 8 \Omega m )

I = J ⋅ A = J A cos ( θ ) 5 = J ( 100 4 ) × cos ( 60 ) J = 5 × 50 = 250 A m − 2 Now, E = ρ × J = 44 × 1 0 − 8 × 250 = 11 × 1 0 − 5 V m − 1

Start Free Mock LoginAlready registered? Login here

Physics - Current Electricity Question with Solution | TestHub

PhysicsCurrent ElectricityMicroscopicEasy2 minQB

PhysicsEasysingle choice

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 $6 0^{\circ}$ with the unit vector of area of cross-section.

The magnitude of electric field at every point of the conductor is: (resistivity of magnesium $ρ = 44 \times 1 0^{- 8} \Omega m$ )

Options:

Answer:

Solution:

$I = J \cdot A = J A cos (θ)$

$5 = J (\frac{4}{100}) \times cos (60)$

$J = 5 \times 50 = 250 A m^{- 2}$

Now, $E = ρ \times J$

$= 44 \times 1 0^{- 8} \times 250 = 11 \times 1 0^{- 5} V m^{- 1}$

Stream:JEESubject:PhysicsTopic:Current ElectricitySubtopic:Microscopic

⏱ 2mℹ️ Source: QB

Doubts & Discussion

Loading discussions...