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.)

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An object of mass 2.0 kg is attached to the top of a vertical spring that is anchored to the floor. The unstressed length of the

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Answer:

The value is $A = 0.014 \ m$

Explanation:

From the question we are told that

The mass of the object is $m = 2.0 \ kg$

The unstressed length of the string is $l = 0.08 \ m$

The length of the spring when it is at equilibrium is $l_e = 5.9 \ cm = 0.059 \ m$

The initial speed (maximum speed)of the spring when given a downward blow $v = 0.30 \ m/s$

Generally the maximum speed of the spring is mathematically represented as

$u = A * w$

Here A is maximum height above the floor (i.e the maximum amplitude)

and $w$ is the angular frequency which is mathematically represented as

$w = \sqrt{\frac{k}{m} }$

$u = A * \sqrt{\frac{k}{m} }$

=> $A = u * \sqrt{\frac{m}{k} }$

Gnerally the length of the compression(Here an assumption that the spring was compressed to the ground by the hammer is made) by the hammer is mathematically represented as

$b = l -l_e$