JEE Mains · Physics · STD 12 - Current electricity
- JEE Mains 2026
- Easy
- Numerical
A cylindrical conductor of length 2m and area of cross-section \( 0.2\,mm^{2} \) carries an electric current of 1.6 A when its ends are connected to a 2V battery. Mobility of electrons in the conductor is \( \alpha\times10^{-3} m^{2}/V.s \). The value of \( \alpha \) is:
(electron concentration \( =5\times10^{28}/m^{3} \) and electron charge \( =1.6\times10^{-19} C \))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) 1
Step-by-step Solution
Detailed explanation
\(V _{\ell}=\mu E =\mu \times \frac{ V }{\ell}\)
\(I = neAV _{ d }\)
\(V_d=\frac{I}{n e A}\)
\(\mu=\frac{ I \ell}{\text { NneA }}\)
\(\mu=\frac{1.6 \times 3}{2 \times 5 \times 10^{26} \times 1.6 \times 10^{-19} \times 2 \times 10^{-7}}\)
\(\mu=1 \times 10^{-3} m^2 / v \cdot s\)
\(\alpha=1\)
\(I = neAV _{ d }\)
\(V_d=\frac{I}{n e A}\)
\(\mu=\frac{ I \ell}{\text { NneA }}\)
\(\mu=\frac{1.6 \times 3}{2 \times 5 \times 10^{26} \times 1.6 \times 10^{-19} \times 2 \times 10^{-7}}\)
\(\mu=1 \times 10^{-3} m^2 / v \cdot s\)
\(\alpha=1\)
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