Question

A total charge of 9.0 mC passes through a cross-sectional area of a nichrome wire in 3.6s. The number of electrons passing through the cross-sectional area in 10s is

Answer 1

Given :

• total charge = 9.0 mC = 0.009 C

Each electron has a charge of :

$1.6 \times 10 {}^{ - 19} \: C$

For producing 1 Cuolomb charge we need :

• $\mathrm{\dfrac{1}{1.6 \times 10 {}^{ - 19} } }$

• $\dfrac{10 {}^{19} }{1.6}$

• $\dfrac{10\times 10 {}^{19} }{16}$

• $\dfrac{100 \times 10 {}^{18} }{16}$

• $\mathrm{6.24 \times 10 {}^{18} \: \: electrons}$

Now, for producing 0.009 C of charge, the number of electrons required is :

• $0.009 \times 6.24 \times {10}^{18}$

• $0.05616 \times 10 {}^{18}$

• $\mathrm{5.616 \times 10 {}^{16} \: \: electons}$

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So, Number of electrons passing through the cross section in 3.6 seconds is :

$\mathrm{5.616 \times 10 {}^{16} \: \: electrons}$

Number of electrons passing through it in 1 Second is :

• $\dfrac{5.616 \times {10}^{16} }{3.6}$

• $\mathrm{1.56 \times 10 {}^{16} \: \: electrons}$

Now, in 10 seconds the number of electrons passing through it is :

• $10 \times \mathrm{1.56 \times 10 {}^{16} \: \: }$

• $\mathrm{1.56 \times 10 {}^{17} \: \: electrons}$

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$\mathrm{ \#TeeNForeveR}$