Question

A copper wire of cross sectional area $3m{m}^2$ carrying a current of $4A$ has ${10}^{29}$ free electrons/$m^3$. If this wire is now placed in a field of induction $0.15T$ perpendicular to wire. The force on each electron is :

• A. $20\times {10}^{-25}N$
• B. $37.5\times {10}^{-25}N$
• C. $10\times {10}^{-25}N$
• D. $41.7\times {10}^{-25}N$

Answer 1

The correct option is A $20\times {10}^{-25}N$

The current in a conductor is given by

$i=nqAv$
where $v$ is drift velocity, $A$ is cross sectional area and $n$ is number of free electrons
$v=\frac{i}{nqA}$
Force on each electron$F=qvB$
$=q\frac{i}{nqA}B$
$=\frac{Bi}{nA}$
$=\frac{0.15\times 4}{{10}^{29}\times 3\times {10}^{-6}}$
$=20\times {10}^{-25}N$