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Q.

An electron of energy $1800 e V$ describes a circular path in magnetic field of flux density $0.4 T$ . The radius of path is $(q = 1.6 \times 10^{- 19} C, m_{e} = 9.1 \times 10^{- 31} k g)$

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a

$2.58 \times 10^{- 4} m$

b

$3.58 \times 10^{- 4} m$

c

$2.58 \times 10^{- 3} m$

d

$3.58 \times 10^{- 3} m$

answer is B.

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Detailed Solution

$E = \frac{1}{2} m ν^{2} \Rightarrow ν = \sqrt{\frac{2 E}{m}} (∴ m_{e} = m) r = \frac{m ν}{B e} = \frac{m}{B e} \sqrt{\frac{2 E}{m}} = \frac{\sqrt{2 m E}}{B e} r = \frac{\sqrt{2 \times 1800 \times 1.6 \times 10^{- 19} \times 9.1 \times 10^{- 31}}}{1.6 \times 10^{- 19} \times 0.4} = 3.58 \times 10^{- 4} m$

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