Gujarati
1. Electric Charges and Fields
medium

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

A

$10^{-8}$

B

$10^{-7}$

C

$10^{-6}$

D

$10^{-5}$

(KVPY-2018)

Solution

(d)

If $Q$ is charge contained in $L$ length of beam of area $A$, then

$L \times A \times \rho=Q$

where, $\rho=$ charge density of beam.

So,

$\rho=\frac{Q}{L \times A}=\frac{Q / t}{L / t \times A}=\frac{I}{v \times A}$

$=\frac{500 \times 10^{-6}}{3 \times 10^7 \times 150 \times 10^{-6}}$

$=\frac{5}{3 \times 1.5} \times 10^{-5}=1.1 \times 10^{-5} \,Cm ^{-3}$

Standard 12
Physics

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