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3 years ago

How can electricty be used to create a magnetic field

2 answers:

5 0

A magnetic field pulls and pushes electrons in some objects near them to make them move metals like copper have electrons that are moved easily and can be readily moved from their orbits if a magnet is moved quickly through coil of copper wire electrons move and electricity is made u can use this but maybe u will want to shorten it down

Strike441 [17]3 years ago

4 0

When a charged particle such as an electron, proton or ion is in motion, magnetic lines of force rotate around the particle. Since electrical current moving through a wire consists of electrons in motion, there is a magnetic field around the wire. i hope u have smileing day

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QUICK ITS DUE IN 6 MINUTES. create a scenario that applies the 3 laws of motion. Explain in complete sentences how each law if d

yKpoI14uk [10]

Answer:

In the first law, an object will not change its motion unless a force acts on it. In the second law, the force on an object is equal to its mass times its acceleration. In the third law, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

5 0

2 years ago

A circular loop carrying a current of 1.6 A is oriented in a magnetic field of 0.30 T. The loop has an area of 0.14 m 2 and is m

konstantin123 [22]

Answer:

The torque on the loop is $2.4 \times 10^{-2}$ Nm

Explanation:

Given:

Current $I = 1.6$ A

Magnetic field $B = 0.30$ T

Area of loop $A = 0.14$ $m^{2}$

Angle between magnetic field and area vector $\theta =$ 21°

Form the formula of torque in case of magnetic field,

г $= MB \sin \theta$

Where $M =$ magnetic moment

$M = IA$

г $= IAB \sin 21$

г $= 1.6 \times 0.30 \times 0.14 \times 0.3583$

г $=2.4 \times 10^{-2}$ Nm

Therefore, the torque on the loop is $2.4 \times 10^{-2}$ Nm

7 0

3 years ago

the moon revolves around the earth in a nearly circular orbit kept by gravitational force exerted by the earth work done will be

rodikova [14]

Answer:

Zero because the applied force is perpendicular to the motion of the object.

No work is done on an object moving is a circular path about a central attractive force.

Any work done in such a case would result in a change in the orbit.

3 0

2 years ago

A 10 gauge copper wire carries a current of 23 A. Assuming one free electron per copper atom, calculate the magnitude of the dri

Reptile [31]

Question:

A 10 gauge copper wire carries a current of 15 A. Assuming one free electron per copper atom, calculate the drift velocity of the electrons. (The cross-sectional area of a 10-gauge wire is 5.261 mm².)

Answer:

3.22 x 10⁻⁴ m/s

Explanation:

The drift velocity (v) of the electrons in a wire (copper wire in this case) carrying current (I) is given by;

v = $\frac{I}{nqA}$

Where;

n = number of free electrons per cubic meter

q = electron charge

A = cross-sectional area of the wire

First let's calculate the number of free electrons per cubic meter (n)

Known constants:

density of copper, ρ = 8.95 x 10³kg/m³

molar mass of copper, M = 63.5 x 10⁻³kg/mol

Avogadro's number, Nₐ = 6.02 x 10²³ particles/mol

But;

The number of copper atoms, N, per cubic meter is given by;

N = (Nₐ x ρ / M) -------------(ii)

Substitute the values of Nₐ, ρ and M into equation (ii) as follows;

N = (6.02 x 10²³ x 8.95 x 10³) / 63.5 x 10⁻³

N = 8.49 x 10²⁸ atom/m³

Since there is one free electron per copper atom, the number of free electrons per cubic meter is simply;

n = 8.49 x 10²⁸ electrons/m³

Now let's calculate the drift electron

Known values from question:

A = 5.261 mm² = 5.261 x 10⁻⁶m²

I = 23A

q = 1.6 x 10⁻¹⁹C

Substitute these values into equation (i) as follows;

v = $\frac{I}{nqA}$

v = $\frac{23}{8.49*10^{28} * 1.6 *10^{-19} * 5.261*10^{-6}}$

v = 3.22 x 10⁻⁴ m/s

Therefore, the drift electron is 3.22 x 10⁻⁴ m/s

6 0

3 years ago

I need help ASAP 8th GRADE SCIENCE

vaieri [72.5K]

Answer:

The answer is gravitational force

Explanation:

4 0

2 years ago

Read 2 more answers