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11.Dual Nature of Radiation and matter
medium
A narrow electron beam passes undeviated through an electric field $E = 3 \times {10^4}volt/m$ and an overlapping magnetic field $B = 2 \times {10^{ - 3}}Weber/{m^2}$. If electric field and magnetic field are mutually perpendicular. The speed of the electrons is
A
$60 m/s$
B
$10.3 \times {10^7}m/s$
C
$1.5 \times {10^7}m/s$
D
$0.67 \times {10^{ - 7}}m/s$
Solution
(c) $eE = evB \Rightarrow v = \frac{E}{B} = \frac{{3 \times {{10}^4}}}{{2 \times {{10}^{ – 3}}}} = 1.5 \times {10^7}m/s$
Standard 12
Physics
