Question

A long, straight wire lies along the z-axis and carries a 4.20-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500 mm segment of the wire centered at the origin.

Answer 1

Complete question:

A long, straight wire lies along the z-axis and carries a 4.00-A current in the +z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.500-mm segment of the wire centered at the origin:

(a) x = 2.00 m, y = 0, z = 0;

(b) x = 0, y = 2.00 m, z = 0;

(c) x = 2.00 m, y = 2.00 m, z = 0; (d) x = 0, y = 0, z = 2.00 m.

Answer:

a) $5.25*10^-1^1 T$;

b) $5.25*10^-^1^1 T$;

c) $1.86*10^-^1^1 T$;

d) 0

Explanation:

Given

I = 4.20A

P = 0.500mm

a) r= (2.00m) i

∆$T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2$;

$B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (2.00)^3$;

$(B= 5.25*10^-^1^1T) j$

b) r = (2.00m) j

∆$T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2$;

$B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (2.00)^3$;

$(B= 5.25*10^-^1^1T) i$

c) r = (2.00m) (i+j)

$(i/j) r = \sqrt{2} (2.00m)$;

∆$T*r = (0.500*10^-^3)(2.00)= 1.00*10^-^3m^2$;

$B = [(1.00*10^-^7 T.m/A)(4.20A)(1*10^-^3m^2)] / (\sqrt{2})(2.00)^3$;

$(B= 1.86*10^-^1^1T)(i-j)$

d) r = (2.00m) k

∆$T*r = (0.500*10^-^3)(2.00) k*k = 0$;

B = 0

Answer 2

The Biot-Savart law and the solution of the determinate allow to find the results for the magnetic field produced at various points by a current segment in the z direction are:

       a) B = 5.25 10-11 j^    T

       b) B = 5.25 10-11 i^   T

       c) B = 1.856 10-11 (-i^ + j^ )    T

        d) B = 0   T

Given parameters

• The value of the current in the + z direction is: I = 4.20 A

• The length of the current segment is l = 0.500mm = 5 10-4 m in the origin position.  

To find

• Magnetic field

        a) r = 2.00 i^ m

        b) r = 2.00 j^ m

        c) r = (2.00 i^ + 2.00 j^  ) m

        d) r = 2 k^ m

The Biot-Savart law gives the value of the magnetic field produced at a point by a segment of current at a given position.

        B = $\frac{\mu_o }{4\pi } \ I \ \frac{dl x r^ }{r^2}$  

where μ₀ is the permeability of the vacuum. I the current, dl a vector in the direction of the current, r  el unitary vector and r² the distance.

The best method to find the magnetic field is to solve for the determinant.

         $B = \frac{\mu_o}{4\pi } \ I \ \left[\begin{array}{ccc}i&j&k\\dl_x&dl_y&dl_z\\r_x&r_y&r_z\end{array}\right]$  

Let's use this expression for each case:

 a)   Let's find the distance using Pythagoras' theorem.

       r = $\sqrt{(x-x_o)^2 +(y-y_o)^2 + (z-z_o)^2}$

It indicates that the current segment is at the origin of the system, therefore its coordinates are zero.

       r = 2.00 m

Let's calculate.

     $B= \frac{4 \pi \ 10^{-7} }{4\pi } \ \frac{4.20}{2^2} \ \left[\begin{array}{ccc}i&j&k\\0&0&5\\1&0&0\end{array}\right] \ 10^{-4}$

     B = 5.25 10⁻¹¹ j

b) The distance is a scalar therefore it has the same value, We look for the magnetic field.

    $B= 10^{-7} \ \frac{4.2}{4} \ \left[\begin{array}{ccc}i&j&k\\0&0&5\\0&1&0\end{array}\right] 10^{-4}$

   B = 5.25 10-11 i

c)  Let's look for the distance.

      r = $\sqrt{2^2 +2^2}$  

      r = 2 √2 m

The unit vector is:

      r ^ =$\frac{1}{\sqrt{2} } \ ( i + j)$ )

Let's calculate the magnetic field.

      $B= 10^{-7 } \ \frac{4.2}{8} \ \frac{1}{\sqrt{2} } \ \left[\begin{array}{ccc}i&j&k\\0&0&5\\1&1&0\end{array}\right] 10^{-5}$

     B = 1.856 10⁻¹¹ (-i + j)

d) Let's calculatethe magnetic field.

     $B= 10^{-7} \frac{4.2}{4} \ \left[\begin{array}{ccc}i&j&0\\0&0&5\\0&0&1\end{array}\right] 10^{-4}$

 

     B = 0  

In conclusion using the Biot-Savart law and the determinate solution we can find the results for the magnetic field produced at various points by a current segment in the z direction are:

       a) B = 5.25 10⁻¹¹ j^     Y

       b) B = 5.25 10⁻¹¹ i^    T

       c) B = 1.856 10⁻¹¹ (-i^ + j^ )   T

       d) B = 0  T

Learn more here:  brainly.com/question/1121860