A particle having a charge 100 e is revolving | vedclass

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(A) The magnetic field $B$ at the center of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Here,the charge $q = 100e$ is re

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$10^{-17} \mu_0$

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(A) The magnetic field $B$ at the center of a circular loop carrying current $I$ is given by $B = \frac{\mu_0 I}{2r}$.Here,the charge $q = 100e$ is revolving with a frequency $f = 1 \ r.p.s$.The equivalent current $I$ is given by $I = qf = 100e 	imes 1 = 100e \ A$.Given radius $r = 0.8 \ m$ and $e = 1.6 	imes 10^{-19} \ C$.Substituting these values into the formula:$B = \frac{\mu_0 	imes 100e}{2 	imes 0.8} = \frac{\mu_0 	imes 100 	imes 1.6 	imes 10^{-19}}{1.6}$.$B = \frac{\mu_0 	imes 100 	imes 1.6 	imes 10^{-19}}{1.6} = 100 	imes 10^{-19} \mu_0 = 10^{-17} \mu_0 \ T$.

$A$ particle having a charge $100 e$ is revolving in a circular path of radius $0.8 \ m$ with $1 \ r.p.s$. The magnetic field produced at the center of the circle in $SI$ unit is ($\mu_0$ is the permeability of vacuum,$e = 1.6 \times 10^{-19} \ C$).

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