Question

An electron beam carries a current of 5 microampere calculate number of electron passing through a point per second and magnetic field produced at a distance of 50 centimetre

Answer 1

Explanation:

q=it

q=5×10^-6×1

q=5×10^-6

now, for no. of electrons we use

q= ne

5×10^-6= n×1.6×19^-19

n=3.125×10^13 Ans.

now magnetic field produce ,

B=u•/4π×i/r

B= 10^-7× 5×10^-6/50×10^-2

B= 10^-7×10^-5

B=10^-12 Ans.

Answer 2

Number of electrons are $3.125\times 10^{13}$.

Magnetic field produced at a distance of 50 cm is $2\times 10^{-12}\ T$.

Explanation:

Given that,

Current in an electron beam,

$I=5\ \mu A\\\\I=5\times 10^{-6}\ A$

(a) Let there are n number of electrons. Using quantization of charge as :

$I=\dfrac{q}{t}\\\\I=\dfrac{ne}{t}\\\\n=\dfrac{It}{e}\\\\n=\dfrac{5\times 10^{-6}\times 1}{1.6\times 10^{-19}}\\\\n=3.125\times 10^{13}$

The number of electrons are $3.125\times 10^{13}$.

(b) Magnetic field produced at a distance of 50 cm is given by :

$B=\dfrac{\mu_o I}{2\pi r}\\\\B=\dfrac{4\pi \times 10^{-7}\times 5\times 10^{-6} }{2\pi \times 0.5}\\\\B=2\times 10^{-12}\ T$

Hence, this is the required solution.

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