An iron ring of relative permeability has windings of insulated copper wire
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Answer:Given,
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
Explanation:
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