Question

A long, straight wire of radius 5.0 mm carries a current of 20A. i) Calculate the magnetic field at the surface of the wire, and ii) calculate the perpendicular distance, from the axis of the wire, at which the magnitude of magnetic field will be half of its value at the wire surface.

Answer 1

Answer on Question #74424 - Physics - Electromagnetism

Question:

A long, straight wire of radius $5.0 \mathrm{~mm}$ carries a current of 20A. i) Calculate the magnetic field at the surface of the wire, and ii) calculate the perpendicular distance, from the axis of the wire, at which the magnitude of magnetic field will be half of its value at the wire surface.

Solution:

We use Ampère's circuit law: the integrated magnetic field around a closed loop

$$\oint \left(\vec {B}, \vec {d l}\right) = \mu_ {0} I,$$

where $I$ -- the electric current through the loop, $\mu_0\approx 1.2566\times 10^{-6}(\mathrm{N / A^2})$ -- magnetic constant in vacuum.

i). If the loop is a circle with radius $R_0 = 5(mm) = 5 \times 10^{-3}(m)$ , then $\vec{B}$ will be directed along the loop and $B_0 = |\vec{B}| = const$ on the circle loop. Therefore

$$\oint \left(\vec {B}, \vec {d l}\right) = 2 \pi R _ {0} B _ {0},$$

and the magnitude of magnetic field

$$B _ {0} = \frac {\mu_ {0} I}{2 \pi R _ {0}} \approx \frac {1 . 2 5 6 6 \times 1 0 ^ {- 6} \times 2 0}{2 \times 3 . 1 4 \times 5 \times 1 0 ^ {- 3}} \approx 8 \times 1 0 ^ {- 4} (T)$$

ii). Due Ampère's circuit law, for a circle loop around the wire, if the radius of the loop $R$ is greater than the radius of the wire $R_0$ , the magnitude of magnetic field is:

$$B = \frac {\mu_ {0} I}{2 \pi R}$$

or

$$R = \frac {\mu_ {0} I}{2 \pi B} = \frac {\mu_ {0} I}{2 \pi B _ {0}} \frac {B _ {0}}{R} = \frac {B _ {0} R _ {0}}{B}.$$

And if $B = B_0 / 2$ then

$$R = \frac {B _ {0} R _ {0}}{B} = \frac {B _ {0} R _ {0}}{B _ {0} / 2} = 2 R _ {0} = 2 \times 5 \times 1 0 ^ {- 3} = 1 0 ^ {- 2} (m)$$

Fig. The wire (blue), the magnetic field at the surface of the wire (red), and the magnitude of magnetic field for the distance $R = 2R_{0}$ (green).

Answer:

i). the magnetic field at the surface of the wire: $B_0 \approx 8 \times 10^{-4}(T)$

ii). the perpendicular distance from the axis of the wire, at which the magnitude of magnetic field will be half of its value at the wire surface: $R = 10^{-2}(m)$

Comments:

1. Ampère's circuital law – for example:

https://en.wikipedia.org/wiki/Amp%C3%A8re%27s_circuital_law

(the table -- "Forms of the original circuital law written in SI units")

2. we use standard SI units:

T—Tesla, m—metre, N—Newton, A -- Ampere

https://en.wikipedia.org/wiki/International_System_of_Units

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