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A short current element dl⃗ = (0.500 mm)j^ carries a current of 5.40 A in the same direction as dl⃗ . Point P is located at r⃗ = ( -0.730 m)i^+ (0.390m)k^. Part A Find the magnetic field at P produced by this current element. Enter the x, y, and z components of the magnetic field separated by commas.

Answer :

Answer:

The magnetic field along x axis is

[tex]B_{x}=1.670\times10^{-10}\ T[/tex]

The magnetic field along y axis is zero.

The magnetic field along z axis is

[tex]B_{z}=3.484\times10^{-10}\ T[/tex]

Explanation:

Given that,

Length of the current element [tex]dl=(0.5\times10^{-3})j[/tex]

Current in y direction = 5.40 A

Point P located at [tex]\vec{r}=(-0.730)i+(0.390)k[/tex]

The distance is

[tex]|\vec{r}|=\sqrt{(0.730)^2+(0.390)^2}[/tex]

[tex]|\vec{r}|=0.827\ m[/tex]

We need to calculate the magnetic field

Using Biot-savart law

[tex]B=\dfrac{\mu_{0}}{4\pi}\timesI\times\dfrac{\vec{dl}\times\vec{r}}{|\vec{r}|^3}[/tex]

Put the value into the formula

[tex]B=10^{-7}\times5.40\times\dfrac{(0.5\times10^{-3})\times(-0.730)i+(0.390)k}{(0.827)^3}[/tex]

We need to calculate the value of [tex]\vec{dl}\times\vec{r}[/tex]

[tex]\vec{dl}\times\vec{r}=(0.5\times10^{-3})\times(-0.730)i+(0.390)k[/tex]

[tex]\vec{dl}\times\vec{r}=i(0.350\times0.5\times10^{-3}-0)+k(0+0.730\times0.5\times10^{-3})[/tex]

[tex]\vec{dl}\times\vec{r}=0.000175i+0.000365k[/tex]

Put the value into the formula of magnetic field

[tex]B=10^{-7}\times5.40\times\dfrac{(0.000175i+0.000365k)}{(0.827)^3}[/tex]

[tex]B=1.670\times10^{-10}i+3.484\times10^{-10}k[/tex]

Hence, The magnetic field along x axis is

[tex]B_{x}=1.670\times10^{-10}\ T[/tex]

The magnetic field along y axis is zero.

The magnetic field along z axis is

[tex]B_{z}=3.484\times10^{-10}\ T[/tex]

The Biot-Savart law and the determinant resolution allow to find the result for the magnetic field produced by a current segment is:

  • The magnetic field is:  (1,670 i^  + 3,484 k^ ) 10⁻¹⁰ T

Given parameters

  • The current element length dl = 0.5 10-3 j m
  • The current   i = 5.40 A
  • The point   r = (-0.730 i + 0.390 j) m

To find

  • Magnetic field

The Biot-Savart law indicates the value of the magnetic field at a point produced by a circulating current  

           [tex]B= \frac{\mu_o}{4\pi }\ i \ \frac{dl \ x\ r}{r^3}[/tex]

where B is the magnetic field, uo the permeability of the vacuum, i the current of the vector in the direction of the current, r the distance to the point of interest.

Let's find the distance using the Pythagorean theorem.

            [tex]r= \sqrt{x^2 +y^2} \\r= \sqrt{0.730^2 + 0.390^2}[/tex]  

            r = 0.827 m

One of the best methods of finding the magnetic field is to solve for the determinant.

          [tex]B= \frac{\mu_o}{4\pi } \ \frac{i}{r^3} \ \left[\begin{array}{ccc}i&j&k\\l_x&l_y&l_z\\r_x&r_y&r_z\end{array}\right][/tex]  

let's calculate.

           [tex]B= 10^{-7} \ \frac{5.40}{0.827^3} \ \left[\begin{array}{ccc}i&j&k\\0&0.5 \ 10^{-3}&0\\-0.730&0&0.390\end{array}\right][/tex]

           B = 1,670 10⁻¹⁰ i + 3,484 10⁻¹⁰ k

In conclusion, using the Biot-Savart law and the determinant resolution we can find the result for the magnetic field produced by a current segment is:

  • The magnetic field is: (1,670 i^ + 3,484 k^ ) 10⁻¹⁰ T

Learn more here: brainly.com/question/12604783

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