Physics · Moving Charges And Magnetism · NEET
For a long straight wire the field falls as 1/r, not 1/r². So B = μ₀I/(2πr). If you double the distance r, the field becomes half, not one-quarter. The 1/r² (inverse square) law belongs to a single point charge (Coulomb) or the far-axis field of a dipole, not to a long straight wire. NEET loves to punish students who mix these up.
Use B = μ₀I/(2πr) for a long STRAIGHT wire at perpendicular distance r. Use B = μ₀I/(2R) for the CENTRE of a full CIRCULAR loop of radius R. They look similar but one has 2π and the other has just 2. A quick check: straight wire has the π; the loop centre does not.
Use the right-hand grip rule. Grip the wire with your right hand so the thumb points along the conventional current I. Your curled fingers then point along the circular field lines. Above a wire carrying current to the right, B points out of the page on one side and into the page on the other. On the axis passing through the wire itself, the field is zero.
For the field OUTSIDE the wire, no. As long as you are outside the wire, it behaves like all the current is concentrated on the axis, so B = μ₀I/(2πr) uses only I and r. Thickness matters only INSIDE a thick wire (that is a separate Ampere's-law result: B = μ₀Ir/2πa²).
μ₀ is the permeability of free space. Its value is 4π × 10⁻⁷ T·m/A. Whenever you see μ₀/(2π) it becomes 2 × 10⁻⁷, which makes numerical work fast: B = 2 × 10⁻⁷ × I/r tesla.
An infinitely long straight conductor carries a current of 5 A. An electron moves with speed 10⁵ m/s parallel to the conductor. The perpendicular distance between the electron and the conductor is 20 cm at an instant. The magnitude of the force experienced by the electron at that instant is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B = μ₀I/(2πr), where I is the current and r is the perpendicular distance from the wire. With μ₀ = 4π×10⁻⁷, this simplifies to B = 2×10⁻⁷ × I/r tesla.
They form concentric circles around the wire. Use the right-hand grip rule: thumb along current, curled fingers give the field direction.
Because B ∝ 1/r, doubling r halves the field. Tripling r makes it one-third. This is a linear inverse, not inverse-square.
The clean B = μ₀I/(2πr) result holds exactly only for an infinite (or very long) straight wire. For a finite or semi-infinite wire you must use the Biot–Savart form with sinθ terms, giving a smaller field.
On the wire's own axis the external field expression gives 0 as r→0 in idealised sense, and the field of one wire can be cancelled by another wire's field at a neutral point where the two contributions are equal and opposite.
