If resistivity of copper conductor is 1.7 × 10-8 Ω m and electric field is 100 Vm -1, then current density will be

Question

If resistivity of copper conductor is 1.7 × 10-8 Ω m and electric field is 100 Vm -1, then current density will be (A) 6 × 109 Am -2 (B) 1.7 × 10-6 Am -2 (C) 1.7 × 10-10 Am -2 (D) 6 × 107 Am -2

Tardigrade Admin · Accepted Answer

Current density, J=σ E where, σ= conductivity =(1/ text resistivity )=(1/ρ)=(1/1.7 × 10-8 Ω m ) ⇒ J=(1/1.7 × 10-8) × 100=(100/17) × 109=6 × 109 Am -2

Q. If resistivity of copper conductor is $1.7 \times 1 0^{- 8} Ω m$ and electric field is $100 V m^{- 1}$ , then current density will be

$6 \times 1 0^{9} A m^{- 2}$

$1.7 \times 1 0^{- 6} A m^{- 2}$

$1.7 \times 1 0^{- 10} A m^{- 2}$

$6 \times 1 0^{7} A m^{- 2}$

Current density, $J = σ E$
where, $σ =$ conductivity $= \frac{1}{resistivity} = \frac{1}{ρ} = \frac{1}{1.7 \times 1 0 ^{- 8} Ω m}$
$\Rightarrow J = \frac{1}{1.7 \times 1 0 ^{- 8}} \times 100 = \frac{100}{17} \times 1 0^{9} = 6 \times 1 0^{9} A m^{- 2}$

Q. If resistivity of copper conductor is 1.7×10−8Ωm and electric field is 100Vm−1, then current density will be

6×109Am−2

1.7×10−6Am−2

1.7×10−10Am−2

6×107Am−2