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2 years ago

A source produces 20 crests and 20 troughs in 4 seconds. The second crest is 3 cm away from the first crest.Calculate :

Physics

1 answer:

sineoko [7]2 years ago

3 0

Answer:

Solution given:

No of waves[N] =20crests & 20 troughs

=20waves

Time[T]=4seconds

distance[d]=3cm=0.03m

Now

Wave length=3cm=3 × ${10}^{ - 2}m$

Frequency= $\frac{No of waves}{time}$

= $\frac{20}{4}$ =5Hertz

and

Wave speed:wave length×frequency=3 × ${10}^{ - 2}m$ ×5=1.5 × ${10}^{ - 1} \tt{ {ms}^{ - 1}}$ .

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Answer:

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Explanation:

First, look at the picture to understand the problem before to solve it.

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Here, the point is located inside the cilinder, just between the wire and the inner layer of the conductor. Therefore, we only consider the wire's current to calculate the magnetic field as follows:

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$B=\frac{u_{0}I_{w}}{2\pi d_{1}} =\frac{52 \times4\pi \times10^{-7} }{2\pi 1.1 \times 10^{-3}} =9.45 \times10^{-3} T\\$

μ0 is the constant of proportionality

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b. d2=3.6 mm

Here, the point is located in the surface of the cilinder. Therefore, we have to consider the current density of the conductor to calculate the magnetic field as follows:

J: current density

c: outer radius

b: inner radius

The cilinder's current is negative, as it goes on opposite direction than the wire's current.

$J= \frac {-I_{c}}{\pi(c^{2}-b^{2} ) }}$

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$B=\frac{u_{0}(I_{w}-JA_{s})}{2\pi d_{2} } \\A_{s}=\pi (d_{2}^{2}-b^2)=4.40\times10^{-6} m^2\\$

$B=\frac{6.68\times10^{-5}}{8.14\times10^{-5}} =0.820 T$

c. d3=7.4 mm

Here, the point is located out of the cilinder. Therefore, we have to consider both, the conductor's current and the wire's current as follows:

$B=\frac{u_{0}(I_w-I_c)}{2\pi d_3 } =\frac{2.011\times10^-5}{3.441\times10^{-4}} =0.0584 T$

As we see, the magnitud of the magnetic field is greater inside the conductor, because of the density of current and the material's nature.

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