Question

A long, straight wire lies along the zz-axis and carries a 3.90-AA current in the z z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.600 mmmm segment of the wire centered at the origin.

Answer 1

The question is incomplete. The complete question is :

A long, straight wire lies along the z-axis and carries a 3.90-A current in the + z-direction. Find the magnetic field (magnitude and direction) produced at the following points by a 0.600 mm segment of the wire centered at the origin.

A) x=2.00m,y=0, z=0

Bx,By,Bz = ? T

Enter your answers numerically separated by commas.

B) x=0, y=2.00m, z=0

C) x=2.00m, y=2.00m, z=0

D) x=0, y=0, z=2.00m

Solution :

The expression of the magnetic field using the Biot Savart's law is given by :

`$d \vec B=\frac{\mu_0 I\vec{dl} * \vec r}{4 \pi r^3}$`

a). The position vector is on the positive x direction.

`$\vec r = (2 \ m) \ \hat i$`

`$|r| = 2 \ m$`

The magnetic field is

`$d \vec B=\frac{\mu_0 I\vec{dl} * \vec r}{4 \pi r^3}$`

`$d \vec B=(4 \pi * 10^(-7) * 3.9 * 0.6 * 10^(-3) *\hat k * (2 ) \hat i )/(4 \pi * (2)^3)$`

`$d \vec B=(5.85 * 10^(-11) \ T)\hat j$`

The magnetic field is
`$(0, \ 5.85 * 10^(-11) \ T, \ 0).$`

b). The position vector is in the positive y-direction.

`$\vec r = (2 \ m) \ \hat j$`

`$|r| = 2 \ m$`

The magnetic field is

`$d \vec B=\frac{\mu_0 I\vec{dl} * \vec r}{4 \pi r^3}$`

`$d \vec B=(4 \pi * 10^(-7) * 3.9 * 0.6 * 10^(-3) *\hat k * (2 ) \hat j )/(4 \pi * (2)^3)$`

`$d \vec B=(5.85 * 10^(-11) \ T)(-\hat{i})$`

The magnetic field is
`$(- 5.85 * 10^(-11) \ T, \ 0, \ 0).$`

c). The position vector is :

`$\vec r = (2)\hat i + (2)\hat j$`

`$|\vec r| = √((2)^2+(2)^2)$`

`$=2.828 \ m$`

The magnetic field is

`$d \vec B=\frac{\mu_0 I\vec{dl} * \vec r}{4 \pi r^3}$`

`$d \vec B=(4 \pi * 10^(-7) * 3.9 * 0.6 * 10^(-3) *\hat k * ((2)\hat i + (2) \hat j) )/(4 \pi * (2.828)^3)$`

`$=(4.13* 10^(-11))\hat j+(4.13* 10^(-11))(-\hat i)$`

The magnitude of the magnetic field is :

`$|d\vec B|=\sqrt{(4.13* 10^(-11))^2+(4.13* 10^(-11))^2}$`

`$=5.84 * 10^(-11) \ T$`

Therefore, the magnetic field is
`$(-4.13 * 10^(-11)\ T, \ 4.13 * 10^(-11)\ T, \ 0 )$`

d). The position vector is in the positive y-direction.

`$\vec r = (2 \ m) \ \hat k$`

`$|r| = 2 \ m$`

The magnetic field is

`$d \vec B=\frac{\mu_0 I\vec{dl} * \vec r}{4 \pi r^3}$`

`$d \vec B=(4 \pi * 10^(-7) * 3.9 * 0.6 * 10^(-3) *\hat k * (2 ) \hat k )/(4 \pi * (2)^3)$`

= 0 T

The magnetic field is (0, 0, 0)