What is the magnetic field outside a long straight current carrying wire equation
The correct answer and explanation is:
Correct Answer:
The magnetic field BB outside a long straight current-carrying wire is given by the equation: B=μ0I2πrB = \frac{\mu_0 I}{2 \pi r}
Where:
- BB is the magnetic field strength
- μ0\mu_0 is the permeability of free space (4π×10−7 T\cdotpm/A)(4\pi \times 10^{-7} \ \text{T·m/A})
- II is the current flowing through the wire
- rr is the radial distance from the center of the wire
Explanation:
This formula originates from Ampère’s Law, a fundamental principle in electromagnetism. Ampère’s Law relates the integrated magnetic field around a closed loop to the electric current passing through the loop. In mathematical terms, it is written as: ∮B⃗⋅dl⃗=μ0Ienclosed\oint \vec{B} \cdot d\vec{l} = \mu_0 I_{\text{enclosed}}
To apply this law to a long straight wire, consider a circular loop of radius rr centered around the wire. Due to symmetry, the magnetic field at every point on this loop has the same magnitude and is tangential to the loop. The path integral then becomes: B⋅(2πr)=μ0IB \cdot (2\pi r) = \mu_0 I
Solving for BB gives: B=μ0I2πrB = \frac{\mu_0 I}{2\pi r}
This equation shows that the magnetic field decreases with distance from the wire. The field lines form concentric circles around the wire, and their density decreases as one moves further away, indicating the weakening of the field.
This behavior is crucial in electrical engineering and physics, especially when analyzing the forces between current-carrying wires, the operation of magnetic sensors, and the design of devices like inductors and transformers. The field exists only outside the wire in this equation; a different expression applies inside the wire if the wire has finite thickness and uniform current distribution.