Question

An electron moves in a circular path of radius $15 \,cm$ in a magnetic field of intensity $ B=4\times {{10}^{-4}}\,T $ . The velocity of electron is :

• A. $ 1.05\times {{10}^{7}}m/s $
• B. $ 5\times {{10}^{6}}m/s $
• C. $ 3.2\times {{10}^{3}}m/s $
• D. zero

Answer 1

The Correct Option is A

Magnetic force provides- the necessary centripetal force.
The magnetic force acting on the electron is
$ F=qvB $ ... (i)
Centripetal force is $ F=\frac{m{{v}^{2}}}{r} $ ... (ii)
Equating Eqs. (i) and (ii), we get
$ r=\frac{mv}{eB} $
$ \Rightarrow $ $ v=\frac{eBr}{m} $
Given, $ B=4\times {{10}^{-4}}T$,
$e=1.6\times {{10}^{-19}}C $ ,
$ r=0.15\,m$,
$m=9.1\times {{10}^{-31}}kg $
$ \therefore $ $ v=\frac{1.6\times {{10}^{-19}}\times 4\times {{10}^{-4}}\times 0.15}{9.1\times {{10}^{-31}}} $
$ v=1.05\times {{10}^{7}}m/s $