Question

Using Ampere's circuital law, obtain the
expression for the magnetic field due to a long
solenoid at a point inside the solenoid on its
axis.​

Answer 1

Answer:

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Answer 2

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=∫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

o

(nLI)

⟹ B=μ

o

nI

solution