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3 years ago

10 points

Physics

1 answer:

zysi [14]3 years ago

3 0

Answer: inside nucleus= 84 outside nucleus= 36

Explanation:

Recall that an atom is made up of three subatomic particles named protons, neutrons and electrons. The proton and neutrons are found inside the nucleus while electrons are found outside the atom nucleus.

So, given that:

Atomic number of krypton = 36

Mass number of krypton = 84

The number of protons = 36 (same as atomic number)

The number of neutrons = (Mass number - Number of protons) 84 - 36 = 48

Thus, particles inside nucleus = (protons + neutrons) = 84

While particles (electrons) outside nucleus = 36 (same as the atomic number which is 36).

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2 years ago

a solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of

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Complete question:

A solenoid that is 98.6 cm long has a cross-sectional area of 24.3 cm2. There are 1310 turns of a wire carrying a current of 6.75 A. (a) Calculate the energy density of the magnetic field inside the solenoid. (b) Find the total energy stored in the magnetic field there (neglect end effects).

Answer:

(a) the energy density of the magnetic field inside the solenoid is 50.53 J/m³

(b) the total energy stored in the magnetic field is 0.121 J

Explanation:

Given;

length of the solenoid, L = 98.6 cm = 0.986 m

cross-sectional area of the solenoid, A = 24.3 cm² = 24.3 x 10⁻⁴ m²

number of turns of the solenoid, N = 1310 turns

The magnitude of the magnetic field inside the solenoid is given by;

B = μ₀nI

B = μ₀(N/L)I

Where;

μ₀ is permeability of free space, = 4π x 10⁻⁷ m/A

$B = \frac{4\pi*10^{-7}*1310*6.75}{0.986} \\\\B = 0.01127 \ T$

(a) Calculate the energy density of the magnetic field inside the solenoid

$u = \frac{B^2}{2 \mu_o}\\\\u = \frac{(0.01127)^2}{2*4\pi *10^{-7}} \\\\u = 50.53 \ J/m^3$

(b) Find the total energy stored in the magnetic field

U = uV

U = u (AL)

U = 50.53 (24.3 x 10⁻⁴ x 0.986)

U = 0.121 J

8 0

3 years ago