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Jlenok [28]
3 years ago
9

Where in our daily lives do we encounter "background" radiation​

Physics
1 answer:
Tcecarenko [31]3 years ago
8 0

Answer:

Explanation:Naturally occurring radioactive materials are present in its crust, the floors and walls of our homes, schools, or offices and in the food we eat and drink. There are radioactive gases in the air we breathe. Our own bodies - muscles, bones, and tissue - contain naturally occurring radioactive elements.

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A hollow cylinder with an inner radius of 5 mm and an outer radius of 26 mm conducts a 4-A current flowing parallel to the axis
bearhunter [10]

Answer:

B = 38.2μT

Explanation:

By the Ampere's law you have that the magnetic field generated by a current, in a wire, is given by:

B=\frac{\mu_o I_r}{2\pi r}     (1)

μo: magnetic permeability of vacuum = 4π*10^-7 T/A

r: distance from the center of the cylinder, in which B is calculated

Ir: current for the distance r

In this case, you first calculate the current Ir, by using the following relation:

I_r=JA_r

J: current density

Ar: cross sectional area for r in the hollow cylinder

Ar is given by  A_r=\pi(r^2-R_1^2)

The current density is given by the total area and the total current:

J=\frac{I_T}{A_T}=\frac{I_T}{\pi(R_2^2-R_1^2)}

R2: outer radius = 26mm = 26*10^-3 m

R1: inner radius = 5 mm = 5*10^-3 m

IT: total current  = 4 A

Then, the current in the wire for a distance r is:

I_r=JA_r=\frac{I_T}{\pi(R_2^2-R_1^2)}\pi(r^2-R_1^2)\\\\I_r=I_T\frac{r^2-R_1^2}{R_2^2-R_1^2}  (2)

You replace the last result of equation (2) into the equation (1):

B=\frac{\mu_oI_T}{2\pi r}(\frac{r^2-R_1^2}{R_2^2-R_1^2})

Finally. you replace the values of all parameters:

B=\frac{(4\pi*10^{-7}T/A)(4A)}{2\PI (12*10^{-3}m)}(\frac{(12*10^{-3})^2-(5*10^{-3}m)^2}{(26*10^{-3}m)^2-(5*10^{-3}m)^2})\\\\B=3.82*10^{-5}T=38.2\mu T

hence, the magnitude of the magnetic field at a point 12 mm from the center of the hollow cylinder, is 38.2μT

8 0
3 years ago
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