Physics · Moving Charges And Magnetism · NEET
Inside, only the current within your Amperian loop of radius r counts. Since current is spread evenly, the enclosed current grows as the area, i.e. I_enc = I*(r^2/R^2). Applying B*(2*pi*r) = mu0*I_enc gives B = mu0*I*r/(2*pi*R^2), so B rises with r. Outside (r > R), the loop already encloses the ENTIRE current I, so I_enc stays fixed at I. Now B*(2*pi*r) = mu0*I gives B = mu0*I/(2*pi*r), which falls as 1/r because the same current is being 'shared' over a bigger loop.
Current density J = I/(pi*R^2) is uniform. The area of your Amperian circle of radius r is pi*r^2. So enclosed current I_enc = J*(pi*r^2) = I*(r^2/R^2). The r^2 comes from AREA, not circumference. This is the single step most students miss - they wrongly use r instead of r^2.
Yes. At r = 0, B = mu0*I*(0)/(2*pi*R^2) = 0. On the axis there is no current enclosed by a zero-radius loop, so B = 0. The field then rises linearly outward. This is a common one-mark trap: field at the axis is zero, not maximum.
At the surface, r = R. Both formulas give the same value there: B_max = mu0*I/(2*pi*R). Just inside (rising line) and just outside (falling 1/r curve) both meet at this peak, so B is continuous at the boundary - no jump.
Only a SOLID wire with uniform current. For a hollow cylinder (current only on the surface or shell), the field INSIDE the hollow region is zero because no current is enclosed. Read the wire type carefully - 'solid', 'thick', 'uniformly distributed over cross-section' all mean use B = mu0*I*r/(2*pi*R^2) inside.
A long straight wire of radius a carries a steady current I. The current is uniformly distributed over its cross-section. The ratio of the magnetic fields B and B' at radial distances a/2 and 2a respectively, from the axis of the wire, is:
From Ampere's circuital law for a long straight wire of circular cross-section carrying a steady current, the variation of magnetic field in the inside and outside region of the wire is:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
B = mu0*I*r/(2*pi*R^2), where r is the distance from the axis (r < R), R is the wire radius, and I is the total current. B is proportional to r, so it is zero at the centre and maximum at the surface.
B = mu0*I/(2*pi*r) for r > R. This is identical to the field of a thin wire, because the Amperian loop encloses the full current I. B falls as 1/r.
At the surface, r = R. There B_max = mu0*I/(2*pi*R). Inside it rises linearly to this value; outside it decays as 1/r.
An Amperian loop of radius r = 0 encloses no current, so B*(2*pi*r) = mu0*I_enc gives B = 0. Physically, on the central axis the field contributions from current on all sides cancel.
For a hollow cylinder carrying current on its outer part, the field in the empty inner region is zero (no enclosed current). The rising-line formula only applies when current is spread through a SOLID cross-section.
