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# A beam of light travelling along $x$ - axis is described by the electric field $E_y = (600 V/m) \sin \omega (t - x/c)$ then maximum magnetic force on a charge $q = 2e$, moving along $y$ - axis with a speed of $3.0 \times 10^7m/s$ is $(e = 1.6 \times 10^{-19} C)$

$\begin {array} {1 1} (a)\;19.2 \times 10^{-17}N & \quad (b)\;1.92 \times 10^{-17}N \\ (c)\;0.192 N & \quad (d)\;\text{None of these} \end {array}$

Maximum magnetic field is given by $B_o = \large\frac{E_o}{c}$
Here, $E_o = 600\: V/m$
So, $B_o = \large\frac{600}{3 \times 10^8}$$= 2 \times 10^{-6}T$
Maximum magnetic force imposed on a given charge is
$F_m = qvB_o = 2evB_o = 2 \times 1.6 \times 10^{-19} \times 3 \times 10^7 \times 2 \times 10^{-6}$
$= 1.92 \times 10^{-17}N$
Ans : (b)