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zaharov [31]
2 years ago
13

A circular coil of five turns and a diameter of 30.0 cm is oriented in a vertical plane with its axis perpendicular to the horiz

ontal component of the Earth's magnetic field. A horizontal compass placed at the coil's center is made to deflect 45.0° from magnetic north by a current of 0.600 A in the coil.(a) What is the horizontal component of the Earth's magnetic field?
Physics
1 answer:
____ [38]2 years ago
5 0

The horizontal component of the magnetic field is 12.6 μT.

The magnetic influence on moving electric currents, electric charges, and magnetic materials is described by a magnetic field, which is a vector field. When a charge moves through a magnetic field, a force that is perpendicular to both its own velocity and the magnetic field operates on it.

The horizontal component of the Earth's magnetic field is perpendicular to the axis of a circular coil with five turns and a diameter of D = 30.0 cm that is vertically orientated.

A coil current of I = 0.600 A causes a horizontal compass to deflect 45.0° from magnetic north when it is positioned in the coil's center.

Let B be the magnetic field and R be the radius of the circular coil.

Then the horizontal component of the Earth's magnetic field is given as:

B(h) = B(coil) = μ₀ NI / 2R

B(h) = (4π × 10⁻⁷ ) (5)(0.6) / 0.3

B(h) = 12.6 μT

Learn more about magnetic field here:

brainly.com/question/14411049

#SPJ4

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In order to find which rational number is between 0 and 1, let's convert them into their decimal form:

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Can you please solve this for me urgently want to make sure if my answers are correct?
MA_775_DIABLO [31]

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\sum d_{x} =  - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\  { \underline{d _{x} =  -  5.702 \: m}} \\  \\  \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0  \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y}  =  - 11.476 \: m}}

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Z. A force that gives a 8-kg objet an acceleration of 1.6 m/s^2 would give a 2-kg object an
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Answer:

\boxed {\boxed {\sf D.\ 6.4\ m/s^2}}

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Substitute these values into the formula.

F= 8 \ kg * 1.6 \ m/s^2

Multiply.

F= 12.8 \ kg*m/s^2

<h3>2. Acceleration of the 2nd Object </h3>

Now,  use the force we just calculated to complete the second part of the problem. We use the same formula:

F= m \times a

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Substitute the values into the formula.

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\frac {12.8 \ kg*m/s^2}{2 \ kg}= \frac{2\ kg* a}{2 \ kg}

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The units of kilograms cancel.

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