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Alja [10]
4 years ago
9

The magnetic force on a wire 274 cm long is . If electrons move through the wire in 1.90 s, what is the magnitude of magnetic fi

eld that is perpendicular to the wire?
Physics
1 answer:
kozerog [31]4 years ago
3 0

Answer:

Matter & Energy

Math Review

Kinematics  

Defining Motion

Graphing Motion

Kinematic Equations

Free Fall

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Dynamics

Newton's 1st Law

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Newton's 2nd Law

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Energy

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Thermal Physics  

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Electrostatics  

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MAGNETISM

Magnetic Fields

The Compass

Electromagnetism  

Electromagnetism

In 1820, Danish physicist Hans Christian Oersted found that a current running through a wire created a magnetic field, kicking off the modern study of electromagnetism.

Moving electric charges create magnetic fields. You can test this by placing a compass near a current-carrying wire. The compass will line up with the induced magnetic field.

To determine the direction of the electrically-induced magnetic field due to a long straight current-carrying wire, use the first right hand rule (RHR) by pointing your right-hand thumb in the direction of positive current flow. The curve of your fingers then shows the direction of the magnetic field around a wire (depicted at right).

You can obtain an even stronger magnetic field by wrapping a coil of wire in a series of loops known as a solenoid and flowing current through the wire. This is known as an electromagnet. You can make the magnetic field from the electromagnet even stronger by placing a piece of iron inside the coils of wire. The second right hand rule tells you the direction of the magnetic field due to an electromagnet. Wrap your fingers around the solenoid in the direction of positive current flow. Your thumb will point toward the north end of the induced magnetic field, as shown below.

Explanation:

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Answer:

66.375 x 10⁻⁶ C/m

Explanation:

Using Gauss's law which states that the net electric flux (∅) through a closed surface is the ratio of the enclosed charge (Q) to the permittivity (ε₀) of the medium. This can be represented as ;

∅ = Q / ε₀        -----------------(i)

Where;

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ε₀ = permittivity of free space (which is air, since it is enclosed in a bag) = 8.85 x 10⁻¹² Nm²/C²

Now, let's first get the charge (Q) by substituting the values above into equation (i) as follows;

7.5 x 10⁵ = Q / (8.85 x 10⁻¹²)

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Now, we can find the linear charge density (L) which is the ratio of the charge(Q) to the length (l) of the rod. i.e

L = Q / l     ----------------------(ii)

Where;

Q = 66.375 x 10⁻⁷ C

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Substitute these values into equation (ii) as follows;

L = 66.375 x 10⁻⁷C / 0.1m

L = 66.375 x 10⁻⁶ C/m

Therefore, the linear charge density (charge per unit length) on the rod is 66.375 x 10⁻⁶ C/m.

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Answer:

<h2><em>6000 counts per second</em></h2>

Explanation:

If a sample emits 2000 counts per second when the detector is 1 meter from the sample, then;

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