Question

A solenoid of length $1.0m$ radius $1cm$ and total turns $1000$ wound on it, carries a current of $5A$. Calculate the magnitude of the axial magnetic field inside the solenoid. If an electron was to move with a speed of $104m{s}^{-}$ along the axis of this current carrying solenoid, what be the force experienced by this electron?

Answer 1

Magnetic field inside a solenoid,
$B={\mu }_{0}nI$
$n=\frac{N}{l}=\frac{1000turns}{1.0m}=1000turns/m$
$I=5A$
$\therefore B=(4\pi \times {10}^{-7})\times 1000\times 5$
$=20\times 3.14\times {10}^{-4}J=6.28\times {10}^{-3}T$, along the axis
Force experienced by electron
$F_m=qvB\sin \theta$
Here $q=-e,v={10}^{4}m/s,$
$\theta =$ angle between $\overrightarrow{v}$ and $\overrightarrow{B}=0$
$\therefore F_m=-evB\sin 0^o=0\left(zero\right)$