An electron gun G emits electrons of energy 2keV travelling in the + ve X-direction. The electrons are required to hit the spot of S where GS = 0.1m, and the line GS makes an angle of 60º v with the x-axis an shown in figure. A uniform magnetic field B parallel to GS exists in the region outside the electron gun. The minimum value of B needed to make the electron hit S is

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$6.0 \times 10^{3} T$

$7.23 \times 10^{- 3} T$

$4.73 \times 10^{- 3} T$

$9 \times 10^{3} T$

answer is B.

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Detailed Solution

$K = 2 Kev = 2 \times 10^{3} \times 1.6 \times 10^{- 19} J Gs = 0.1m (pitch) θ = 60 ° q = 1.6 \times 10^{- 19} c m = {9.1×10}^{- 31} kg Pitch (Gs) = v_{11} T Gs = v \cos θ \frac{2 π m}{Bq} \frac{1}{2} {mv}^{2} = K v = \sqrt{\frac{2K}{m}} \begin{array}{l} Gs = \sqrt{\frac{2K}{m}} \cos θ \frac{2 π m}{Bq} \\ B = \sqrt{\frac{{2Km}^{2}}{m}} (\cos θ) \frac{2 π}{Bq} \\ B = \sqrt{2Km} (\cos 60 °) \frac{2 π}{(Gs) q} \\ B = \sqrt{2Km} (\frac{1}{2}) \frac{2 π}{(Gs) q} \\ B = \sqrt{2Km} \frac{π}{(Gs) q} \\ B = \sqrt{2 \times 2 \times 10^{3} \times 1.6 \times 10^{- 19} \times 9.1 \times 10^{- 31}} \times (\frac{3.14}{(0.1) \times 1.6 \times 10^{- 19}}) \\ B = 4.73 \times 10^{- 3} T \end{array}$