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an electron beam travels with a velocity of $1.6 \times 10^{7}\;ms^{-1}$ perpendicularly to magnetic field of intensity 0.1 T. The radius of the path of electron beam $(m_e=9 \times 10^{-31}kg)$

\[\begin {array} {1 1} (a)\;9 \times 10^{-5}m & \quad (b)\;9 \times 10^{-2}m \\ (c)\;9 \times 10^{-4}m & \quad (d)\;9 \times 10^{-3}m \end {array}\]

1 Answer

$(d)\;9 \times 10^{-3}m $
answered Nov 7, 2013 by pady_1
 

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