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Gnom [1K]
1 year ago
7

A closed curve encircles several conductors. The line integral around this curve is (image attached below)

Physics
1 answer:
Masteriza [31]1 year ago
5 0

The net current in the conductors and the value of the line integral

  • I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A
  • The resultant remains same 3.2 *10^4 Tm

This is further explained below.

<h3>What is the net current in the conductors?</h3>

Generally,

To put it another way, the total current In flowing across a surface S (contained by C) is proportional to the line integral of the magnetic B-field (in tesla, T).

\oint_C \mathbf{B} \cdot \mathrm{d}\boldsymbol{\ell} = \mu_0 \iint_S \mathbf{J} \cdot \mathrm{d}\mathbf{S} = \mu_0I_\mathrm{enc}

I=\frac{3.2\cdot 10^{-4}}{4\pi \cdot 10^{-7}}=254.77\, A

B)

In conclusion, It is possible for the line integral to go around the loop in either direction (clockwise or counterclockwise), the vector area dS to point in either of the two normal directions and Ienc, which is the net current passing through the surface S, to be positive in either direction—but both directions can be chosen as positive in this example. The right-hand rule solves these ambiguities.

The resultant remains the same at 3.2 *10^4 Tm

Read more about conductors

brainly.com/question/8426444

#SPJ1

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Frequencia = 1.25 Hz

Explanation:

<u>Dados los siguientes datos;</u>

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Matemáticamente, la velocidad de una onda viene dada por la fórmula;

Velocidad = Longitud \; de \; onda * Frequencia

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Frequencia = \frac {Velocidad}{Longitud \; de \; onda}

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A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
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Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

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13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

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3 years ago
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