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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class12  >>  Electromagnetic Waves

A light beam travelling in the x-direction is described by the electric field $E_y = 300 \: V/m \: \sin \omega\: (t – x/c)$. An electron is constrained to move along the y-direction with a speed of $2 \times 10^7 m/s$. Find the maximum magnetic force on the electron

$\begin {array} {1 1} (a)\;32 \times 10^{-16} N & \quad (b)\;3.2 \times 10^{-16} N \\ (c)\;32 \times 10^{-18} Nabc & \quad (d)\;3.2 \times 10^{-186} N \end {array}$

 

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1 Answer

$B_o = \large\frac{E_o}{c}$$ = \large\frac{300}{3 \times 10^8}$$ = 100 \times 10^{-8}T$
$Fm = B_oqv = 9.1 \times 10^{-31} \times 100 \times 10^{-8} \times 2 \times 10^7 N = 3.2 \times 10^{-18} N $
Ans : (d)
answered Mar 1, 2014 by thanvigandhi_1
 

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