A charged particle is accelerated through a potential difference of 12kV and acquires a speed of 1⋅0×106ms−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.

V=12kV E=Vl Now, F=qE=qVl or, a=Fm=qVmlu=1×106m/s or u=2×qVml×l=2×qm×12×103 or 1×106=2×qm×12×103⇒1012=24×103×qm⇒mq=24×1031012=24×10−9r=muqB=24×10−9×1×1062×10−1=12×10−2m=12cm

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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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answer is 12.

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Detailed Solution

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