Changing the current changes the "what" of an electromagnet

2 answers:

8 0

You take one spring winding on the one rod like iron rod. this winding iron rod called coil. when you give electric supply to both pole on that rod then that rod or coil produce one electro magnetic field. is called electro magnetism ..... this coil called electromagnet...

Helen [10]3 years ago

4 0

An electromagnet is only magnetic when there's current going through
the wire. If there's no current flowing, then the electromagnet is just a
lump and doesn't attract anything.

The strength of the electromagnet depends on the amount of current
that's flowing through the coil of wire around it.

Changing the current immediately changes ==> the magnitude of the
magnetic field, ==> the strength of the electromagnet, and what it's
able to pick up.

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a thundercloud whose base is 500m above the ground. The potential difference between the base of the cloud and the m ground is 2

kolezko [41]

Answer:

1.6×10⁻⁶ N.

Explanation:

From the question,

F = (V/r)q......................... Equation 1

Where F = Electric force on the raindrop, V = Potential difference between the base of the cloud and the ground, r = distance between the base of the cloud and the ground, q = the charge on a rain drop.

Given: V = 200MV = 200×10⁶ V, r = 500 m, q = 4.0×10⁻¹² C.

Substitute these values into equation 1

F = [(200×10⁶ )/500]×4.0×10⁻¹²

F = 1.6×10⁻⁶ N.

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3 years ago

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frozen [14]

The answer is B........

4 0

4 years ago

Read 2 more answers

1) A spring, which has a spring constant k=7.50 N/m, has been stretched 0.40 m from ts equilibrium position . What the potential

cupoosta [38]

<h3>Answer:</h3>

$\displaystyle U_s = 0.6 \ J$

<h3>General Formulas and Concepts:</h3>

<u>Math</u>

<u>Pre-Algebra</u>

Order of Operations: BPEMDAS

Brackets
Parenthesis
Exponents
Multiplication
Division
Addition
Subtraction

Left to Right<u> </u>

<u>Physics</u>

<u>Energy</u>

Elastic Potential Energy: $\displaystyle U_s = \frac{1}{2} k \triangle x^2$

U is energy (in J)
k is spring constant (in N/m)
Δx is displacement from equilibrium (in m)

<h3>Explanation:</h3>

<u>Step 1: Define</u>

k = 7.50 N/m

Δx = 0.40 m

<u>Step 2: Find Potential Energy</u>

Substitute in variables [Elastic Potential Energy]: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.40 \ m)^2$
Evaluate exponents: $\displaystyle U_s = \frac{1}{2} (7.50 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = (3.75 \ N/m) (0.16 \ m^2)$
Multiply: $\displaystyle U_s = 0.6 \ J$