Question

The electron revolves 6 x 1018 times per second in a circular orbit. The current in the loop is​

Answer 1

$Freq. \: of \: revolution \: (v) = 6 \times {10}^{15} \: {S}^{ - 1} \\ Charge \: of \: electron \: (q) \: = 1.6 \times {10}^{ - 19} C$

$Current \: in \: loop \:(I) = qv$

$= 1.6 \times {10}^{ - 19} \times 6 \times {10}^{15} \\ = 9.6 \times {10}^{ - 4} \\ = 0.96 \: mA$

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Answer 2

As per the data given in the above question

we have to find the current loop

Given,

• Number of electron ,n= 6×10⁸ times per second
• time, t=1

$Current \: or \: I= \frac{q}{t} ........(1)$

where Q is charge and t is time

we know that ,

$q = n \times e \: \: \: \: ..........(2)$

where n is the frequency of revolution in per second

so, above formula becomes from (1) and (2) equation

$I= \frac{n \times \: e }{t} ........(3)$

charge of an electron ,

${e}^{ - } \: charge =1.6× {10}^{ - 19} coulomb$

Now put the values in equation (3)

$I= \frac{6 \times {10}^{8} \times \: 1.6 \times {10}^{ - 19} }{1}$

$I= \frac{9.6 \times \: {10}^{8 - 19} }{1}$

$I= {9.6 \times \: {10}^{ - 11} } \: ampere$

Hence ,

The current in the loop is 9.6×10-¹¹ ampere .

Project code# SPJ2