The note "Middle C" is known to have a frequency of 261.6 Hz. What would

Physics

1 answer:

Gwar [14]3 years ago

8 0

Answer: frequency = 523.2 Hz

Explanation:

One octave higher = 2f of the last octave.

For instance, if C2 = 4 Hz

A octave higher will be C3 in which its frequency will be 2 × 4Hz = 8Hz

Back to the question.

One octave higher frequency will therefore equal to:

2f = 2 × 261.6 = 523.2 Hz

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For part a)
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In this experiment we will observe the magnetic fields produced by a current carrying wire. A long wire is suspended vertically,

Alisiya [41]

Answer:

See explanation

Explanation:

Solution:-

Electric current produces a magnetic field. This magnetic field can be visualized as a pattern of circular field lines surrounding a wire. One way to explore the direction of a magnetic field is with a compass, as shown by a long straight current-carrying wire in. Hall probes can determine the magnitude of the field. Another version of the right hand rule emerges from this exploration and is valid for any current segment—point the thumb in the direction of the current, and the fingers curl in the direction of the magnetic field loops created by it.

Compasses placed near a long straight current-carrying wire indicate that field lines form circular loops centered on the wire. Right hand rule 2 states that, if the right hand thumb points in the direction of the current, the fingers curl in the direction of the field. This rule is consistent with the field mapped for the long straight wire and is valid for any current segment.

( See attachments )

- The equation for the magnetic field strength - B - (magnitude) produced by a long straight current-carrying wire is given by the Biot Savart Law:

$B = \frac{uo*I}{2\pi *r}$

Where,

I : The current,

r : The shortest distance to the wire,

uo : The permeability of free space. = 4π * 10^-7 T. m/A

- Since the wire is very long, the magnitude of the field depends only on distance from the wire r, not on position along the wire. This is one of the simplest cases to calculate the magnetic field strength - B - from a current.

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$B = \frac{uo*I}{4\pi }*\int\frac{dl xr}{r^2}$