Question

Write Ampere's circuital law. Find the
expression of produced magnetic field at the
centre of a long current carrying solenoid with
its help.
​

Answer 1

Answer:

Ampere's circuital law states that line integral of magnetic field around any closed loop is equal to μ

o

times the electric current flowing through the cross-section area enclosed by that loop.

Mathematically, ∮B.dl=μI

Let the current flowing in the solenoid having number of turns per unit length n be I.

Magnitude of magnetic field inside the solenoid is B while at outside is zero.

Now ∮

loop

B.dl=∫Bab

'L+∫Bbc.L′+∫Bcd.L+∫Bda.L′

The value of first term ∫Bab.L=BL

The second and fourth term are zero because angle between magnetic field and the length loop is 90°

The third term is also zero as the value of magnetic field outside the solenoid is zero.

Total current flowing through the loop I

total =(nL)I

From Ampere's circuital law, we get BL=μo(nLI)

⟹ B=μonI