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12.2: Magnetic Field due to a Thin Straight Wire

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  • [ "article:topic", "authorname:openstax", "Thin straight wire", "Magnetic Field", "license:ccby" ]

    How much current is needed to produce a significant magnetic field, perhaps as strong as Earth’s field? Surveyors will tell you that overhead electric power lines create magnetic fields that interfere with their compass readings. Indeed, when Oersted discovered in 1820 that a current in a wire affected a compass needle, he was not dealing with extremely large currents. How does the shape of wires carrying current affect the shape of the magnetic field created? We noted in Chapter 28 that a current loop created a magnetic field similar to that of a bar magnet, but what about a straight wire? We can use the Biot-Savart law to answer all of these questions, including determining the magnetic field of a long straight wire.

    Figure \(\PageIndex{1}\) shows a section of an infinitely long, straight wire that carries a current I. What is the magnetic field at a point P, located a distance R from the wire?

    This figure shows a section of a thin, straight current-carrying wire. Point P is located at distance R from the center of the wire O and at distance r from the piece of the wire dX. Vector r from the piece of the wire dX to the point P forms an angle theta with the wire.

    Figure \(\PageIndex{1}\): A section of a thin, straight current-carrying wire. The independent variable \(\theta\) has the limits \(\theta_1\) and \(\theta_2\).

    Let’s begin by considering the magnetic field due to the current element \(I \, d\vec{x}\) located at the position x. Using the right-hand rule 1 from the previous chapter, \(d\vec{x} \times \hat{r}\) points out of the page for any element along the wire. At point \(P\), therefore, the magnetic fields due to all current elements have the same direction. This means that we can calculate the net field there by evaluating the scalar sum of the contributions of the elements. With

    \[|d\vec{x} \times \hat{r}| = (dx)(1) \, \sin \, \theta \]

    we have from the Biot-Savart law

    \[B = \dfrac{\mu_0}{4\pi} \int_{wire} \dfrac{I \, \sin \, \theta \, dx}{r^2}. \label{BSLaw}\]

    The wire is symmetrical about point \(O\), so we can set the limits of the integration from zero to infinity and double the answer, rather than integrate from negative infinity to positive infinity. Based on the picture and trigonometry, we can write expressions for \(r\) and \(\sin \, \theta\) in terms of x and R, namely:

    \[r = \sqrt{x^2 + R^2}\]

    \[\sin \, \theta = \dfrac{R}{\sqrt{x^2 + R^2}}.\]

    Substituting these expressions into Equation \ref{BSLaw}, the magnetic field integration becomes

    \[B = \dfrac{\mu_0I}{2\pi} \int_0^{\infty} \dfrac{R \, dx}{(x^2 + R^2)^{3/2}}.\]

    Evaluating the integral yields

    \[B = \dfrac{\mu_0I}{2\pi R} \left[\dfrac{x}{(x^2 + R^2)^{1/2}}\right]_0^{\infty}.\]

    Substituting the limits gives us the solution

    \[\boxed{B = \dfrac{\mu_0 I}{2\pi R}.}\]

    The magnetic field lines of the infinite wire are circular and centered at the wire (Figure \(\PageIndex{2}\)), and they are identical in every plane perpendicular to the wire. Since the field decreases with distance from the wire, the spacing of the field lines must increase correspondingly with distance. The direction of this magnetic field may be found with a second form of the right-hand rule (Figure \(\PageIndex{2}\)). If you hold the wire with your right hand so that your thumb points along the current, then your fingers wrap around the wire in the same sense as \(\vec{B}\).

    This figure demonstrates the right-hand rule. The wire is held with the right hand so that the thumb points along the current. The fingers wrap around the wire in the same sense as the magnetic field.

    Figure \(\PageIndex{2}\):​​​​​​. Some magnetic field lines of an infinite wire. The direction of \(B\) can be found with a form of the right-hand rule.

    The direction of the field lines can be observed experimentally by placing several small compass needles on a circle near the wire, as illustrated in Figure \(\PageIndex{3a}\). When there is no current in the wire, the needles align with Earth’s magnetic field. However, when a large current is sent through the wire, the compass needles all point tangent to the circle. Iron filings sprinkled on a horizontal surface also delineate the field lines, as shown in Figure \(\PageIndex{3b}\).

    Figure A shows a circle formed by the small compass needles aligned with Earth’s magnetic field. Figure B shows that iron filings sprinkled on a horizontal surface around a long wire delineate the field lines.

    Figure \(\PageIndex{3}\): The shape of the magnetic field lines of a long wire can be seen using (a) small compass needles and (b) iron filings.

    Example \(\PageIndex{1}\): Calculating Magnetic Field Due to Three Wires

    Three wires sit at the corners of a square, all carrying currents of 2 amps into the page as shown in Figure \(\PageIndex{4}\)​​​​​​​. Calculate the magnitude of the magnetic field at the other corner of the square, point P, if the length of each side of the square is 1 cm.

    Figure shows three wires I1, I2, and I3 with current flowing into the page. Wires form three corners of a square. The magnetic field is determined at the fourth corner of the square that is labeled P.

    Figure \(\PageIndex{4}\)​​​​​​​: Three wires have current flowing into the page. The magnetic field is determined at the fourth corner of the square.

    Strategy

    The magnetic field due to each wire at the desired point is calculated. The diagonal distance is calculated using the Pythagorean theorem. Next, the direction of each magnetic field’s contribution is determined by drawing a circle centered at the point of the wire and out toward the desired point. The direction of the magnetic field contribution from that wire is tangential to the curve. Lastly, working with these vectors, the resultant is calculated.

    Solution

    Wires 1 and 3 both have the same magnitude of magnetic field contribution at point P:

    \[B_1 = B_3 = \dfrac{\mu_0 I}{2\pi R} = \dfrac{(4\pi \times 10^{-7} T \cdot m/A)(2 \, A)}{2\pi (0.01 \, m)} = 4 \times 10^{-5}T.\]

    Wire 2 has a longer distance and a magnetic field contribution at point P of:

    \[B_2 = \dfrac{\mu_0 I}{2\pi R} = \dfrac{(4\pi \times 10^{-7}T \cdot m/A)(2 \, A)}{2 \pi (0.01414 \, m)} = 3 \times 10^{-5}T.\]

    The vectors for each of these magnetic field contributions are shown.

    Figure shows three wires I1, I2, and I3 with current flowing into the page. Wires form three corners of a square. The magnetic field is determined at the fourth corner of the square that is labeled P. Vector B3 is directed from the point P towards the wire I1. Vector B1 is the continuation of the line from the wire I3 to the point P. Vector B2 lies between vectors B1 and B3.

    The magnetic field in the x-direction has contributions from wire 3 and the x-component of wire 2:

    \[B_{net \, x} = -4 \times 10^{-5}T - 2.83 \times 10^{-5}T \, \cos (45^o) = -6 \times 10^{-5}T.\]

    The y-component is similarly the contributions from wire 1 and the y-component of wire 2:

    \[B_{net \, y} = -4 \times 10^{-5}T - 2.83 \times 10^{-5}T \, \sin (45^o) = -6 \times 10^{-5}T.\]

    Therefore, the net magnetic field is the resultant of these two components:

    \[ \begin{align} B_{net} &= \sqrt{B_{net \, x}^2 + B_{net\, y}} \\[5pt] &= \sqrt{(-6 \times 10^{-5}T)^2 + (-6 \times 10^{-5}T)^2} \\[5pt] &= 8.48 \times 10^{-5} T. \end{align}\]

    Significance

    The geometry in this problem results in the magnetic field contributions in the x- and y-directions having the same magnitude. This is not necessarily the case if the currents were different values or if the wires were located in different positions. Regardless of the numerical results, working on the components of the vectors will yield the resulting magnetic field at the point in need.

    Exercise \(\PageIndex{1}\)

    Using Example \(\PageIndex{1}\)​​​​​​​, keeping the currents the same in wires 1 and 3, what should the current be in wire 2 to counteract the magnetic fields from wires 1 and 3 so that there is no net magnetic field at point \(P\)?

    Solution

    4 amps flowing out of the page

    Contributors

    Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0).