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jonny [76]
3 years ago
12

How many electron flow through a light bulb each second if the current flow through the light bulb 0.75A.The electric charge of

one electron is 1.6 x 10-19C
Physics
1 answer:
castortr0y [4]3 years ago
4 0

Answer:

n=4.68\times 10^{18}

Explanation:

The current through the bulb, I = 0.75 A

We need to find the number of electrons flowing per second. We know that the electric current is given by :

I=\dfrac{ne}{t}\\\\n=\dfrac{It}{e}\\\\n=\dfrac{0.75\times 1}{1.6\times 10^{-19}}\\\\n=4.68\times 10^{18}

So, there are 4.68\times 10^{18} electrons flowing per second.

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You have a 1.8m long copper wire. You want to make an N-turn current loop that generates a 2.0mT magnetic field at the center wh
NikAS [45]

Answer:

The diameter is  0.022m.

Explanation:

The magnetic field B at the center of the coil is given by

(1). B = \dfrac{\mu_0 NI}{d}

where \mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2} is the magnetic constant, I is the current, N number of coils, and d is the diameter of the coil.

Now, if we call L the length of the wire, then it must be true that

\pi dN = L <em>(this says </em>N<em> coil circumferences (</em>c=\pi d<em>) fit into </em>L<em> )</em>

\therefore N = \dfrac{L}{\pi d }

putting this into equation (1) we get:

B = \dfrac{\mu_0 IL }{\pi d^2}

solve for d:

\boxed{d = \sqrt{\dfrac{\mu_0I L }{\pi B} }}

putting in the numerical values

\mu_0 = 1.26*10^{-6} m\:kg\:s^{-2} A^{-2}

I =1.3A

L= 1.8m

B =2.0*10^{-3}T

we get:

d = \sqrt{\dfrac{(1.26*10^{-6})(1.3) *1.8 }{\pi (2.0*10^{-3})} }

\boxed{d = 0.022m}

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