Formulas/physics/Moving Charges/Field Due to Finite Straight Wire

Field Due to Finite Straight Wire

Field at perpendicular distance d from a finite wire. φ₁ and φ₂ are angles subtended at the field point from the perpendicular foot to each end of the wire.

Class 12

Derivation

Setup

A straight wire carries current $I$ . The field point P is at perpendicular distance $d$ from the wire. The wire extends from angle $-\phi_1$ to $+\phi_2$ as seen from P, measured from the perpendicular foot.

Integration

For a small element $dy$ at position $y$ on the wire, $\sin\theta = d/\sqrt{d^2+y^2}$ and $r^2 = d^2+y^2$ . From Biot-Savart:

dB = \frac{\mu_0 I}{4\pi} \frac{d\,dy}{(d^2 + y^2)^{3/2}}

Substituting $y = d\tan\phi$ :

dB = \frac{\mu_0 I}{4\pi d}\cos\phi\,d\phi

Integrating from $-\phi_1$ to $\phi_2$ :

\boxed{B = \frac{\mu_0 I}{4\pi d}\,(\sin\phi_1 + \sin\phi_2)}

All contributions point in the same direction, so the integration is scalar.

Special Cases

Semi-infinite wire ( $\phi_1 = 0°$ , $\phi_2 = 90°$ ):

B = \frac{\mu_0 I}{4\pi d}

Infinite wire ( $\phi_1 = \phi_2 = 90°$ ):

B = \frac{\mu_0 I}{2\pi d}

Note

Both $\phi_1$ and $\phi_2$ are positive angles measured from the perpendicular foot to each end. If P is not between the perpendiculars of the two ends, one angle is negative — adjust the formula accordingly.

← Back to Moving Charges