Question

find the direction of force?

Answer 1

$Given \: charge (q) = 3.2 \times 10^{-19} \:c ,\\Velocity (V) = 10^{6} \: m/s ,\\Magnetic \:field \: intensity (B) = 3\: Wb/m^{2},\\Angle (\theta) =60\degree\\Force = F$

$\underline {\pink {By \: Fleming's \:Left\:Hand\: Rule :}}$

$\boxed { \orange { F = qVB sin\theta }}$

/* Substitute the values */

$F = 3.2 \times 10^{-19} \times 10^{6} \times 3 \times \frac{\sqrt{3}}{2} \\= 1.6\times 3\times 1.732 \times 10^{-13}$

$F = 8.3136 \times 10^{-13} \:N$

Therefore.,

$\red{F}\green { = 8.3136 \times 10^{-13} \:N}$

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