TS EAMCET · Physics · Current Electricity
A conductor of length 1.5 m and area of cross-section \(3 \times 10^{-5} \mathrm{~m}^2\) has electrical resistance of \(15 \Omega\). The current density in the conductor for an electric field of \(21 \mathrm{Vm}^{-1}\) is
- A \(0.7 \times 10^6 \mathrm{Am}^{-2}\)
- B \(0.7 \times 10^{-6} \mathrm{Am}^{-2}\)
- C \(0.7 \times 10^{-5} \mathrm{Am}^{-2}\)
- D \(0.7 \times 10^5 \mathrm{Am}^{-2}\)
Answer & Solution
Correct Answer
(D) \(0.7 \times 10^5 \mathrm{Am}^{-2}\)
Step-by-step Solution
Detailed explanation
\(\mathrm{l}=1.5 \mathrm{~m}, \mathrm{~A}=3 \times 10^{-5} \mathrm{~m}^2, \mathrm{R}=15 \Omega, \mathrm{E}=21 \mathrm{Vm}^{-1}\) The current density is…
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