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A charged particle is accelerated through a potential difference of 12kV and acquires a speed of $1 \cdot 0 \times 10^{6} {ms}^{- 1}$ . It is then injected perpendicularly into a magnetic field of strength 0.2T . Find the radius (in cm) of the circle described by it.

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detailed solution

Correct option is L

$\begin{array}{l} V = 12 k V E = \frac{V}{l} \\ Now, F = qE = \frac{qV}{l} or, a = \frac{F}{m} = \frac{qV}{ml} \\ u = 1 \times 10^{6} m / s \\ or u = \sqrt{2 \times \frac{qV}{ml} \times l} = \sqrt{2 \times \frac{q}{m} \times 12 \times 10^{3}} or 1 \times 10^{6} = \sqrt{2 \times \frac{q}{m} \times 12 \times 10^{3}} \\ \Rightarrow 10^{12} = 24 \times 10^{3} \times \frac{q}{m} \Rightarrow \frac{m}{q} = \frac{24 \times 10^{3}}{10^{12}} = 24 \times 10^{- 9} \\ r = \frac{m u}{qB} = \frac{24 \times 10^{- 9} \times 1 \times 10^{6}}{2 \times 10^{- 1}} = 12 \times 10^{- 2} m = 12 cm \end{array}$

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detailed solution

Correct answer is 12

$\begin{array}{l} V = 12 k V E = \frac{V}{l} \\ Now, F = qE = \frac{qV}{l} or, a = \frac{F}{m} = \frac{qV}{ml} \\ u = 1 \times 10^{6} m / s \\ or u = \sqrt{2 \times \frac{qV}{ml} \times l} = \sqrt{2 \times \frac{q}{m} \times 12 \times 10^{3}} or 1 \times 10^{6} = \sqrt{2 \times \frac{q}{m} \times 12 \times 10^{3}} \\ \Rightarrow 10^{12} = 24 \times 10^{3} \times \frac{q}{m} \Rightarrow \frac{m}{q} = \frac{24 \times 10^{3}}{10^{12}} = 24 \times 10^{- 9} \\ r = \frac{m u}{qB} = \frac{24 \times 10^{- 9} \times 1 \times 10^{6}}{2 \times 10^{- 1}} = 12 \times 10^{- 2} m = 12 cm \end{array}$

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