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3 years ago

How does wire thickness affect the power of the electromagnet?

1 answer:

6 0

It does not depend on the thickness of the wire. Well, not directly.

As steveastrouk wrote, the magnetic field depends on the current through the wire and how often the wire goes around the core - the number of loops.

If you use a thinner wire, you can wrap more turns into a given space. Good for the magnetic field! But...

A thinner wire means more resistance. More turns mean a longer wire and that means even more resistance. More resistance means that a given voltage can drive less current through the wire. Less current means a lower magnetic field.

If and when the two effects cancel each other, well that's what math is for.

(Of course you could increase the voltage to keep the current in the thinner wire the same as in the thicker wire, but that would increase the thermal losses in the wire as well.)

You might be interested in

Assume that block A which has a mass of 30 kg is being pushed to the left with a force of 75 N along a frictionless surface. Wha

Veronika [31]

Answer:

The force of friction acting on block B is approximately 26.7N. Note: this result does not match any value from your multiple choice list. Please see comment at the end of this answer.

Explanation:

The acting force F=75N pushes block A into acceleration to the left. Through a kinetic friction force, block B also accelerates to the left, however, the maximum of the friction force (which is unknown) makes block B accelerate by 0.5 m/s^2 slower than the block A, hence appearing it to accelerate with 0.5 m/s^2 to the right relative to the block A.

To solve this problem, start with setting up the net force equations for both block A and B:

$F_{Anet} = m_A\cdot a_A = F - F_{fr}\\F_{Bnet} = m_B\cdot a_B = F_{fr}$

where forces acting to the left are positive and those acting to the right are negative. The friction force F_fr in the first equation is due to A acting on B and in the second equation due to B acting on A. They are opposite in direction but have the same magnitude (Newton's third law). We also know that B accelerates 0.5 slower than A:

$a_B = a_A-0.5 \frac{m}{s^2}$

Now we can solve the system of 3 equations for a_A, a_B and finally for F_fr:

$30kg\cdot a_A = 75N - F_{fr}\\24kg\cdot a_B = F_{fr}\\a_B= a_A-0.5 \frac{m}{s^2}\\\implies \\a_A=\frac{87}{54}\frac{m}{s^2},\,\,\,a_B=\frac{10}{9}\frac{m}{s^2}\\F_{fr} = 24kg \cdot \frac{10}{9}\frac{m}{s^2}=\frac{80}{3}kg\frac{m}{s^2}\approx 26.7N$

The force of friction acting on block B is approximately 26.7N.

This answer has been verified by multiple people and is correct for the provided values in your question. I recommend double-checking the text of your question for any typos and letting us know in the comments section.

6 0

3 years ago

Read 2 more answers

A child on a swing sweeps out a distance of 45 ft on the first pass. If she is allowed to continue swinging until she stops, an

earnstyle [38]

Answer:

d = 90 ft

Explanation:

Here in each swing the distance sweeps by the swing is half of the initial distance that it will move

So here we can say that total distance in whole motion is given as

$d = 45 + \frac{45}{2} + \frac{45}{4} + \frac{45}{8}..........$

since it is half of the distance that it will move in each swing so it would be a geometric progression with common ratio of 1/2

so sum of such GP is given by the formula

$S = \frac{a}{1 - r}$

$d = \frac{45}{1 - \frac{1}{2}}$

$d = 90 ft$

6 0

3 years ago

In an inkjet printer, letters and images are created by squirting drops of ink horizontally at a sheet of paper from a rapidly m

Serga [27]

Answer:

$q = 6.48 \times 10^{-14} C$

Explanation:

Deflection in the drop is due to electric field force

so we will have

$F = qE$

acceleration of the drop is given as

$a = \frac{qE}{m}$

$a = \frac{q(7.75 \times 10^4)}{1.00 \times 10^{-11}}$

$a = 7.75 \times 10^{15} q$

now we know that time to cross the plates is given as

$t = \frac{D}{v}$

$t = \frac{0.02}{18}$

$t = 1.11 \times 10^{-3} s$

now the deflection is given as

$d = \frac{1}{2}at^2$

$0.310 \times 10^{-3} = \frac{1}{2}(7.75 \times 10^{15} q)(1.11 \times 10^{-3})^2$

$0.310 \times 10^{-3} = 4.78 \times 10^9 q$

$q = 6.48 \times 10^{-14} C$

5 0

2 years ago

Velocity question...is it A B Cor D

Ket [755]

In a constant acceleration of 3m per second, after 10 seconds,

3 x 10 = 30

B. 30m/s is your answer

hope this helps :D

6 0

2 years ago

Read 2 more answers

Write a sentence to describe how an image is formed on retina<br>when looking at distant objects.

beks73 [17]

Answer:

An image is formed on the retina with light rays converging most at the cornea and upon entering and exiting the lens. Rays from the top and bottom of the object are traced and produce an inverted real image on the retina. The distance to the object is drawn smaller than scale

7 0

3 years ago