Question

An iron ring of relative permeability 'µ' has windings of insulated copper wire of n turns per metre. When the current in the windings is I, find the expression for the magnetic field in the ring.

Answer 1

Relative permeability = u then, permeability of iron = μ₀u
number of turns per meter = n
current through it = I

Now, At center of solenoid { if a current carrying conductor tightly wounded over a wire , system named as solenoid } , cut an element of thickness dx at x meter from its center .
so, number of turns in dx element is N = ndx
Now, use Biot savart law,
B = μ₀uNiR²/(R² + x²)^(3/2) , Let R is the radius of circular part
= μ₀unidxR²/(R² + x²)^(3/2)
= μ₀uniR²dx/(R² + x²)^(3/2)
After integration we get, Magnetic field at center when length of wire is infinite
B = μ₀unI

Hence, answer is B = B = μ₀unI