Question

State Ampere’s circuital law. Find magnetic field intensity at a point well

inside a current carrying a straight solenoid.​

Answer 1

Answer:

Ampere's Circuital Law states the relationship between the current and the magnetic field created by it. This law states that the integral of magnetic field density (B) along an imaginary closed path is equal to the product of current enclosed by the path and permeability of the medium.

Explanation:

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

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=∫B

ab

.L+∫B

bc

.L

′

+∫B

cd

.L+∫B

da

.L

′

The value of first term ∫B

ab

.L=BL

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

o

.

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=μ, (nLI)

⟹ B=μ,nl