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Elodia [21]
3 years ago
9

The magnetic field of a long, straight, and closely-wound solenoid, inside the solenoid at a point near the center, is 0.645 T.

What is the magnitude of the magnetic field at this point if the current in the solenoid is increased by a factor of 3.00
Physics
1 answer:
Savatey [412]3 years ago
3 0

Answer:

B'=1.935 T      

Explanation:

Given that

magnetic field ,B= 0.645 T

We know that magnetic filed in the solenoid is given as

B=\mu _0 n\ I

I=Current

n=Number of turn per unit length

μ0 =magnetic permeability

Now when the current increased by 3 factors

I'=3 I

Then the magnetic filed

B'=\mu _0 n\ I'

B'=\mu _0 n\ (3I)

B'=3 B

That is why

B' = 3 x 0.645 T

B'=1.935 T

Therefore the new magnetic filed will be 1.935 T.

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Answer:

Density (φ) = 0,8827 Kg/L

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Explanation:

The density formula: φ = \frac{m}{V}

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = \frac{1,56Kg}{1,767145L} = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) /  φ(w)

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Hope this can help you !!

3 0
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Fifty points please help?
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Tectonic plates are large segments of the earth's crust that move slowly. suppose one such plate has an average speed of 4.8 cm
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Time period of the pendulum is given as

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