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AP Physics 2: Algebra Notes

4.3.2 Field Strength, Current, and Distance

AP Syllabus focus: 'The magnetic field due to a long straight wire is proportional to current and inversely proportional to perpendicular distance from the wire’s axis.'

For a long straight wire, magnetic field strength changes in a simple, predictable way: larger current produces a stronger field, while greater distance from the wire weakens it.

Core relationship

A long straight wire with a steady current produces a magnetic field around it.

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A long, straight current-carrying wire produces magnetic field lines that form concentric circles centered on the wire. The arrows indicate the magnetic-field direction around the wire (the direction is tangential to the circles), reinforcing the circular symmetry assumed in the long-wire model. Source

For this idealized geometry, the magnetic field strength at a point depends on two quantities only: the current in the wire and the perpendicular distance from the wire. More current makes the field stronger, while greater distance makes it weaker. This relationship is central because it allows quick comparisons of field strength without needing a full derivation each time.

Perpendicular distance from the wire’s axis: The shortest distance from the point where the field is measured to the central line of the wire.

The emphasis on perpendicular distance matters. The relevant separation is the shortest distance from the point to the wire’s center line, not some other measured length.

B=μ0I2πr B=\dfrac{\mu_0 I}{2\pi r}

B B = magnetic field strength, in tesla

μ0 \mu_0 = constant for free space, approximately 4π×107 Tm/A4\pi\times10^{-7}\ T\cdot m/A

I I = current in the wire, in ampere

r r = perpendicular distance from the wire's axis, in meter

This equation is the exact free-space form of the syllabus statement. It shows that field strength increases linearly with current and decreases inversely with perpendicular distance.

What the proportionalities mean

Direct proportionality to current

If the distance from the wire stays fixed, the field strength changes by the same factor as the current. Double the current, and the field doubles. Triple the current, and the field triples. Reduce the current by half, and the field is halved. This is called a direct proportionality because BB and II change together in a linear way.

This is especially useful when only the current changes. In that case, the geometry stays the same, so the comparison is simple and mostly conceptual.

Inverse proportionality to distance

If the current stays fixed, the field strength changes inversely with distance from the wire. Double the perpendicular distance, and the field becomes half as large. Move three times farther away, and the field becomes one-third as large. Move to half the original distance, and the field doubles.

This is an inverse relationship, not an inverse-square relationship. That distinction is important. The field weakens with 1/r1/r, so it decreases more gradually than a quantity that follows 1/r21/r^2.

Interpreting distance correctly

Because the equation uses rr, problems often test whether you can identify the correct distance. For a long straight wire, every point that is the same perpendicular distance from the axis has the same field magnitude. The wire’s axis is the imaginary line running through its center.

That means:

  • measure from the wire’s center line, not from its outer edge unless the geometry makes that necessary

  • use the shortest distance to the axis

  • do not use a distance measured along the wire

  • do not use the length of a circular path around the wire

If a point is farther from the wire, the field there is weaker even if the current has not changed. If two points are equally far from the axis, the field magnitudes are equal for the ideal long straight wire model.

Comparing two situations

A very efficient AP Physics method is to compare two magnetic fields by using ratios instead of repeatedly substituting constants. From the main equation,

B2B1=I2I1r1r2\dfrac{B_2}{B_1}=\dfrac{I_2}{I_1}\dfrac{r_1}{r_2}

This form is helpful when both current and distance change at the same time. It keeps the physical meaning clear: increasing current strengthens the field, while increasing distance weakens it.

When using this ratio idea, match the variables carefully. A current increase appears as I2/I1I_2/I_1, while a larger distance reduces the field, so the distance factor appears as r1/r2r_1/r_2.

For a fixed distance, a graph of magnetic field strength vs. current is a straight line through the origin. That reflects the direct proportionality between BB and II.

For a fixed current, a graph of magnetic field strength vs. distance is not a straight line.

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Graph of magnetic field magnitude BB versus radial distance rr from the wire’s axis. Outside the wire (r>Rr>R), the curve is labeled B1/rB\propto 1/r, matching the AP Physics long-wire relationship; the plot also highlights that the behavior differs inside the wire (r<Rr<R), where the field increases approximately linearly with rr for uniform current density. Source

It curves downward because the field follows 1/r1/r. If you instead graph magnetic field strength vs. 1/r1/r, you get a straight-line relationship.

These graph shapes help identify whether data fit the model of a long straight wire. They also help distinguish this relationship from inverse-square behavior.

Limits and common mistakes

The equation is based on an ideal long straight wire. It works best when the point of interest is not near the ends of the wire and when the wire is much longer than the distance to the observation point.

Common mistakes include:

  • treating the distance dependence as 1/r21/r^2 instead of 1/r1/r

  • forgetting that the comparison uses perpendicular distance

  • changing both current and distance but accounting for only one of them

  • assuming the field stays the same when the point moves farther away

  • using the wire’s physical length in the equation, even though it does not appear in the long-wire model

FAQ

Air has magnetic properties extremely close to those of a vacuum under ordinary conditions.

For most AP Physics problems, the difference is so small that using $\mu_0$ gives an excellent approximation. That is why the free-space equation is typically used for wires in air unless a problem clearly says otherwise.

Not usually. The standard equation applies to points outside a long straight wire.

Inside a wire with a roughly uniform current distribution, the field does not keep increasing as $1/r$. Instead, it starts at zero at the center and increases with distance from the center until it reaches the surface. So the outside formula should not be extended all the way to $r=0$.

For points outside a long cylindrical wire, the field depends on the total current enclosed, not on how spread out that current is across the wire’s cross-section.

That is why two wires carrying the same total current can produce the same magnetic field at the same outside distance from their axes, even if their diameters are different.

A simple test is to measure field strength with a magnetic field probe.

  • Keep $r$ fixed and vary $I$ to test whether $B$ is proportional to $I$

  • Keep $I$ fixed and vary $r$ to test whether $B$ is proportional to $1/r$

  • Plot $B$ vs. $I$ for a straight line

  • Plot $B$ vs. $1/r$ for another straight line

Straight-line graphs support the model.

If the current changes with time, the magnetic field also changes with time.

The equation for a long straight wire can still describe the instantaneous field in simple situations, but rapidly changing currents can introduce additional effects associated with changing magnetic fields. In that case, the steady-current model is no longer the whole story, and more advanced analysis is needed.

Practice Questions

A long straight wire carries current II. The current is doubled while the observation point is moved to three times its original perpendicular distance from the wire. How does the magnetic field strength change?

  • Uses BIrB\propto \dfrac{I}{r} or an equivalent relationship. (1)

  • Correctly states that the new field is 23\dfrac{2}{3} of the original field. (1)

A very long straight wire carries a steady current of 6.0 A6.0\ A. Point P is 0.020 m0.020\ m from the wire, and point Q is 0.050 m0.050\ m from the wire.

(a) State the equation for the magnetic field strength around a long straight wire in free space.
(b) Determine the ratio BPBQ\dfrac{B_P}{B_Q}.
(c) The current is then changed to 9.0 A9.0\ A, and the field is measured again at point Q. Determine the ratio of the new field at Q to the original field at P.

(a)

  • States B=μ0I2πrB=\dfrac{\mu_0 I}{2\pi r} or an equivalent equation. (1)

(b)

  • Recognizes that current is the same at P and Q, so BPBQ=rQrP\dfrac{B_P}{B_Q}=\dfrac{r_Q}{r_P}. (1)

  • Correctly substitutes distances and obtains 0.0500.020=2.5\dfrac{0.050}{0.020}=2.5. (1)

(c)

  • Uses a valid ratio method, such as BQ,newBP,old=InewIoldrPrQ\dfrac{B_{Q,new}}{B_{P,old}}=\dfrac{I_{new}}{I_{old}}\dfrac{r_P}{r_Q}. (1)

  • Correctly substitutes values: 9.06.00.0200.050\dfrac{9.0}{6.0}\dfrac{0.020}{0.050}. (1)

  • Obtains 0.600.60. (1)

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