In a particle accelerator,a current of 500 ,m | vedclass

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(D) The current $I$ is related to the charge density $\rho$,cross-sectional area $A$,and velocity $v$ by the formula $I = \rho A v$.Given:$I = 500 \,\

Vedclass · Accepted Answer

$10^{-5}$

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(D) The current $I$ is related to the charge density $ho$,cross-sectional area $A$,and velocity $v$ by the formula $I = ho A v$.Given:$I = 500 \,\mu A = 500 	imes 10^{-6} \,A$$v = 3 	imes 10^7 \,m/s$$A = 1.50 \,mm^2 = 1.50 	imes 10^{-6} \,m^2$Rearranging the formula to solve for charge density $ho$:$ho = \frac{I}{A v}$Substituting the values:$ho = \frac{500 	imes 10^{-6}}{(1.50 	imes 10^{-6}) 	imes (3 	imes 10^7)}$$ho = \frac{500}{1.50 	imes 3 	imes 10^7}$$ho = \frac{500}{4.5 	imes 10^7} \approx 111.11 	imes 10^{-9} \approx 1.11 	imes 10^{-7} \,C/m^3$.Wait,re-evaluating the calculation:$ho = \frac{500 	imes 10^{-6}}{1.5 	imes 10^{-6} 	imes 3 	imes 10^7} = \frac{500}{4.5 	imes 10^7} = \frac{500}{45,000,000} = \frac{5}{450,000} = \frac{1}{90,000} \approx 1.11 	imes 10^{-5} \,C/m^3$.Thus,the value is close to $10^{-5} \,C/m^3$.

In a particle accelerator,a current of $500 \,\mu A$ is carried by a proton beam in which each proton has a speed of $3 \times 10^7 \,m/s$. The cross-sectional area of the beam is $1.50 \,mm^2$. The charge density in this beam (in $C/m^3$) is close to

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