# How would you calculate the magnetic field in the following places: 33cm to the north of the wire? 230mm to the south of the wire? 880μm below the wire? 0.18m above the wire?

## A wire is traveling east to west through iron. A 940mA current is running through the wire going towards the east.

Aug 10, 2017

In the order asked: $B = 5.7 \times {10}^{-} 7 \text{T out of the plane, " 8.17xx10^-7"T into the plane, " 2.14xx10^-4"T into the plane, "1.04xx10^-6"T out of the plane}$

#### Explanation:

The magnetic field generated by a current-carrying wire is given by:

$B = \frac{{\mu}_{o} I}{2 \pi r}$

where $I$ is the current through the wire, $r$ is the distance from the wire to the specified point, and ${\mu}_{o}$ is a constant for the permeability of free space

Right hand rule to find direction of magnetic field: We are given the following information:

• $\mapsto I = 940 \times {10}^{-} 3 \text{A}$
• |->mu_o=4pixx10^-7("T"*"m")/"A"
• $\mapsto {r}_{1} = 0.33 \text{m (north)}$
• $\mapsto {r}_{2} = 0.230 \text{m (south)}$
• $\mapsto {r}_{3} = 880 \times {10}^{-} 6 \text{m (south)}$
• $\mapsto {r}_{4} = 0.18 \text{m (north)}$

For the first position, we have:

$B = \frac{{\mu}_{o} I}{2 \pi {r}_{1}}$

=>=((4pixx10^-7("T"*"m")/"A")(940xx10^-3"A"))/(2pi(0.33"m"))

$\implies = 5.7 \times {10}^{-} 7 \text{T}$

By the right hand rule, for a point above the wire, the magnetic field is pointing out of the plane, toward you.

$\therefore B = 5.7 \times {10}^{-} 7 \text{T}$ out of the plane

For the second position, we have:

$B = \frac{{\mu}_{o} I}{2 \pi {r}_{2}}$

=>=((4pixx10^-7("T"*"m")/"A")(940xx10^-3"A"))/(2pi(0.230"m"))

$\implies = 8.17 \times {10}^{-} 7 \text{T}$

By the right hand rule, for a point below the wire, the magnetic field is pointing into the plane, away from you.

$\therefore B = 8.17 \times {10}^{-} 7 \text{T}$ into the plane

For the third position, we have:

$B = \frac{{\mu}_{o} I}{2 \pi {r}_{3}}$

=>=((4pixx10^-7("T"*"m")/"A")(940xx10^-3"A"))/(2pi(880xx10^-6"m"))

$\implies = 2.14 \times {10}^{-} 4 \text{T}$

$\therefore B = 2.14 \times {10}^{-} 4 \text{T}$ into the plane

For the fourth position, we have:

$B = \frac{{\mu}_{o} I}{2 \pi {r}_{4}}$

=>=((4pixx10^-7("T"*"m")/"A")(940xx10^-3"A"))/(2pi(0.18"m"))

$\implies = 1.04 \times {10}^{-} 6 \text{T}$

$\therefore B = 1.04 \times {10}^{-} 6 \text{T}$ out of the plane