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sergeinik [125]
3 years ago
10

An ideal solenoid 20 cm long is wound with 5000 turns of very thin wire. What strength magnetic field is produced at the center

of the solenoid when a current of 10 A flows through it?
Physics
1 answer:
KIM [24]3 years ago
7 0
<h2>Answer:</h2>

0.31425 Tesla

<h2>Explanation:</h2>

The magnetic field strength of a solenoid can be found by using the Ampere's law as follows;

BL = μ₀ x N x I      -------------------(i)

Where;

B = magnetic field strength

L = length of the solenoid

μ₀ = magnetic constant = 1.257 x 10⁻⁶H/m

N = number of turns in the coil of the solenoid

I = current flowing through the coil of the solenoid.

<em>From the question, </em>

L = 20cm = 0.2m

N = 5000 turns

I = 10A

<em>Substitute these values into equation (i) as follows;</em>

B x 0.2 = 1.257 x 10⁻⁶ x 5000 x 10

0.2B = 6.285 x 10⁻²

<em>Solve for B;</em>

B = 6.285 x 10⁻² / 0.2

B = 31.425 x 10⁻²

B = 0.31425 T

Therefore, the magnetic field strength is 0.31425 Tesla

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Answer with Explanation:

We are given that

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0=19-9.8t

9.8t=19

t=\frac{19}{9.8}=1.94 s

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Using the formula

s=19(1.94)-\frac{1}{2}(9.8)(1.94)^2

s=18.4 m

a.The ball rise upto height 18.4 m

b.It take 1.94 s to reach its highest point.

c.Initial velocity=0,s=18.4 m

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18.4=4.9t^2

t^2=\frac{18.4}{4.9}

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Answer:

a) F=1.044\times 10^9\ N

b)F'=1.044\times 10^9\ N

c) F_p=1.0672\times10^{-7}\ N

d) Treat the humans as though they were points or uniform-density spheres.

Explanation:

Given:

  • mass of Mars, M=6.4\times 10^{23}\ kg
  • radius of the Mars, r=3.4\times 10^{6}\ m
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a)

Gravitation force exerted by the Mars on the human body:

F=G.\frac{M.m}{r^2}

where:

G=6.67 \times 10^{-11}\ m^3.kg^{-1}.s^{-2} = gravitational constant

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The magnitude of the gravitational force exerted by the human on Mars is equal to the force by the Mars on human.

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d)

The gravitational constant is a universal value and it remains constant in the Universe and does not depends on the size of the mass.

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