An electron beam travels with a velocity of 1 | vedclass

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(C) When a charged particle moves perpendicularly to a magnetic field,it follows a circular path.The radius $r$ of the circular path is given by the f

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$9 	imes 10^{-4} \ m$

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(C) When a charged particle moves perpendicularly to a magnetic field,it follows a circular path.The radius $r$ of the circular path is given by the formula:$r = \frac{mv}{qB}$Given values:$m = 9 	imes 10^{-31} \ kg$$v = 1.6 	imes 10^7 \ m/s$$q = e = 1.6 	imes 10^{-19} \ C$$B = 0.1 \ T$Substituting these values into the formula:$r = \frac{9 	imes 10^{-31} 	imes 1.6 	imes 10^7}{1.6 	imes 10^{-19} 	imes 0.1}$$r = \frac{9 	imes 1.6 	imes 10^{-24}}{0.16 	imes 10^{-19}}$$r = \frac{14.4 	imes 10^{-24}}{1.6 	imes 10^{-20}}$$r = 9 	imes 10^{-4} \ m$

An electron beam travels with a velocity of $1.6 \times 10^7 \ m/s$ perpendicularly to a magnetic field of intensity $0.1 \ T$. Calculate the radius of the path of the electron beam. (Given: $m_e = 9 \times 10^{-31} \ kg$,$e = 1.6 \times 10^{-19} \ C$)

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