Question

In an expt, a beam of electron passes undeviated through mutually perpendicular elec. and mag fields of respective strength $E=7.2\times {10}^{6}N{C}^{-1}$ and B=2.4 T. The velocity of the electron is

• A. $17.3\times {10}^{7}m{s}^{{-}^1}$
• B. $3\times {10}^{6}m{s}^{{-}^1}$
• C. $2\times {10}^{6}m{s}^{{-}^1}$
• D. $6\times {10}^{6}m{s}^{{-}^1}$

Answer 1

The correct option is B $3\times {10}^{6}m{s}^{{-}^1}$

Force due to the magnetic field and electric field must balance each other

$F_E=qE=eE$

$F_B=q\vartheta B=e\vartheta B$

$\therefore eE=e\vartheta B$

$\therefore \vartheta =\frac{E}{B}=\frac{7.2\times {10}^6}{2.4}$

$=3\times {10}^{6}m{s}^{-1}$