Magnetic field from a finite straight current-carrying wire segment at a distance d is given by which expression?

Magnetic field from a finite straight current-carrying wire segment at a distance d is given by which expression?

• A. B = μ0 I /(2π d) (cos θ1 − cos θ2)
• B. B = μ0 I /(4π d) (cos θ1 − cos θ2)
• C. B = μ0 I /(4π d) (cos θ1 + cos θ2)
• D. B = μ0 I /(4π d) (cos θ2 − cos θ1)

Answer: (B)

The essential idea is to use the Biot–Savart law and integrate along the finite straight wire. Each little segment dl carries current I and contributes a small magnetic field that, when you sum all along the wire, adds up to a simple result at a point a distance d away from the wire. Because the observation point is at a fixed perpendicular distance d, the geometry depends only on how the lines from the observation point to the wire’s ends subtend angles with the wire’s axis. Denoting those end angles by θ1 and θ2, the integration over the wire gives a magnitude B = μ0 I /(4π d) times (cos θ1 − cos θ2). The cos terms come from the way the contributions add up with the varying direction along the wire, and the difference reflects the finite length. This expression reduces to the well-known infinite-wire result in the appropriate limit: as the ends recede to infinity, the angles arrange so that (cos θ1 − cos θ2) becomes 2, giving B = μ0 I /(2π d), the field of an infinite straight wire. The direction of B is tangential around the wire, given by the right-hand rule.

The essential idea is to use the Biot–Savart law and integrate along the finite straight wire. Each little segment dl carries current I and contributes a small magnetic field that, when you sum all along the wire, adds up to a simple result at a point a distance d away from the wire.

Because the observation point is at a fixed perpendicular distance d, the geometry depends only on how the lines from the observation point to the wire’s ends subtend angles with the wire’s axis. Denoting those end angles by θ1 and θ2, the integration over the wire gives a magnitude B = μ0 I /(4π d) times (cos θ1 − cos θ2). The cos terms come from the way the contributions add up with the varying direction along the wire, and the difference reflects the finite length.

This expression reduces to the well-known infinite-wire result in the appropriate limit: as the ends recede to infinity, the angles arrange so that (cos θ1 − cos θ2) becomes 2, giving B = μ0 I /(2π d), the field of an infinite straight wire. The direction of B is tangential around the wire, given by the right-hand rule.