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(A) The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by $T = \frac{2\pi m}{qB}$.For a long solenoid,the magne

Vedclass · Accepted Answer

$250$

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(A) The time period $T$ of a charged particle moving in a uniform magnetic field $B$ is given by $T = \frac{2\pi m}{qB}$.For a long solenoid,the magnetic field inside is $B = \mu_0 n I$,where $n$ is the number of turns per metre and $I$ is the current.Substituting $B$ into the time period formula: $T = \frac{2\pi m}{q \mu_0 n I}$.Rearranging for $n$: $n = \frac{2\pi m}{q \mu_0 I T}$.Given values: $m = 9 	imes 10^{-31} 	ext{ kg}$,$q = 1.6 	imes 10^{-19} 	ext{ C}$,$I = 1.5 	ext{ A}$,$T = 75 	imes 10^{-9} 	ext{ s}$,and $\mu_0 = 4\pi 	imes 10^{-7} 	ext{ N/A}^2$.Substituting these values:$n = \frac{2\pi 	imes 9 	imes 10^{-31}}{1.6 	imes 10^{-19} 	imes 4\pi 	imes 10^{-7} 	imes 1.5 	imes 75 	imes 10^{-9}}$.$n = \frac{18\pi 	imes 10^{-31}}{9.6\pi 	imes 10^{-19} 	imes 10^{-7} 	imes 10^{-9} 	imes 75}$.$n = \frac{18 	imes 10^{-31}}{9.6 	imes 75 	imes 10^{-35}} = \frac{18 	imes 10^4}{720} = \frac{180000}{720} = 250 	ext{ turns/m}$.

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