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Jlenok [28]
3 years ago
15

What would happen if you use a thicker wire around the iron nail of an electromagnet? (thats the whole question)

Physics
1 answer:
puteri [66]3 years ago
3 0

Answer:

When we have a current I, we will have a magnetic field perpendicular to this current.

Then if we have a wire in a "spring" form. then we will have a magnetic field along the center of this "spring".

Now suppose we put an iron object in the middle (where the magnetic field is) then we will magnetize the iron object.

Of course, the intensity of the magnetic field is proportional to the current, given by:

B = (μ*I)/(2*π*r)

Where:

μ is a constant, I is the current and r is the distance between to the current.

Now remember that for a resistor:

R = ρ*L/A

R is the resistance, ρ is the resistivity, which depends on the material of the wire, L is the length of the wire, and A is the cross-section of the wire.

If we increase the area of the wire (if we use a thicker wire).

And the relation between resistance and current is:

I = V/R

Where V is the voltaje.

Now, if we use a thicker wire, then the cross-section area of the wire increases.

Notice in the resistance equation, that the cross-section area is on the denominator, then if we increase the area A, the resistance decreases.

And the resistance is on the denominator of the current equation, then if we decrease R, the current increases.

If the current increases, the magnetic field increases, which means that we will have a stronger electromagnet.

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Which type of matter is gold an element used in making jewelry A.) A heterogeneous mixture B.) A pure substance C.) A crystallin
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A

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How are frequency and wave period related?
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Is there a "real" simple machine that has an efficiency of 100%?
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Answer:

NO

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8 0
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A ski gondola is connected to the top of a hill by a steel cable of length 620 m and diameter 1.5 cm. As the gondola comes to th
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Answer:

(a) 89 m/s

(b) 11000 N

Explanation:

Note that answers are given to 2 significant figures which is what we have in the values in the question.

(a) Speed is given by the ratio of distance to time. In the question, the time given was the time it took the pulse to travel the length of the cable twice. Thus, the distance travelled is twice the length of the cable.

v=\dfrac{2\times 620 \text{ m}}{14\text{ s}} = \dfrac{1240\text{ m}}{14\text{ s}}=88.571428\ldots \text{ m/s}= 89\text{ m/s}

(b) The tension, T, is given by

v =\sqrt{\dfrac{T}{\mu}}

where v is the speed, T is the tension and \mu is the mass per unit length.

Hence,

T = \mu\cdot v^{2}

To determine \mu, we need to know the mass of the cable. We use the density formula:

\rho = \dfrac{m}{V}

where m is the mass and V is the volume.

m=\rho\cdot V

If the length is denoted by l, then

\mu = \dfrac{m}{l} = \dfrac{\rho\cdot V}{l}

T = \dfrac{\rho\cdot V}{l} v^{2}

The density of steel = 8050 kg/m3

The cable is approximately a cylinder with diameter 1.5 cm and length or height of 620 m. Its volume is

V = \pi \dfrac{d^{2}}{4} l

T = \dfrac{\rho\cdot\pi d^2 l}{4l}v^2 = \dfrac{\rho\cdot\pi d^2}{4}v^2

T = \dfrac{8050\times\pi\times0.015^2}{4} \times 88.57^2

T = 11159.4186\ldots \text{ N} = 11000 \text{ N}

4 0
3 years ago
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