A very long straight wire of radius R = 0.003 m carries current 15 A uniformly across its circular cross-section, directed out of the page. Use μ₀ = 4π × 10⁻⁷ T m A⁻¹ and neglect end effects. Apply Ampère's law to derive B(r) for r R. Calculate the field at R/2, R and 2R, identify the maximum, and state its circulation direction.

Question

A very long straight wire of radius R = 0.003 m carries current 15 A uniformly across its circular cross-section, directed out of the page. Use μ₀ = 4π × 10⁻⁷ T m A⁻¹ and neglect end effects. Apply Ampère's law to derive B(r) for r < R and r > R. Calculate the field at R/2, R and 2R, identify the maximum, and state its circulation direction.

Accepted Answer

Inside B = μ₀Ir/(2πR²); outside B = μ₀I/(2πr). B(R/2) = B(2R) = 5.000 × 10^-4 T; maximum B(R) = 0.001 T. Counterclockwise. Cylindrical symmetry makes the field tangent to a circle centred on the wire and constant in magnitude around it. Ampère's law reduces to B(2πr) = μ₀I_enclosed. Uniform current density gives I_enclosed = Iπr²/(πR²) = Ir²/R² for r < R. Therefore B = μ₀Ir/(2πR²), increasing linearly from zero at the centre. For r ≥ R the loop encloses all the current, so B = μ₀I/(2πr). This expression agrees with the internal expression at the surface and then decreases with radius. At R/2 and 2R the magnitude is 5.000 × 10^-4 T. At R the magnitude is 0.001 T, the maximum because the radial dependence changes from increasing to decreasing there. The right-hand grip rule gives counterclockwise circulation for current out of the page. Both expressions have units tesla, and continuity at the surface checks the enclosed-current calculation.

Magnetic field inside and outside a wire · set 10

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