Question

Derive the epression for magnetic field inside a very long solenoid having n turns per unit length carrying a current i

Answer 1

n(pie)/t pie is the si unit of magnetic flux

Answer 2

Magnetic field due to long solenoid is B = μ•ni

•Solenoid : A wire wounded in cylindrical form which behaves as bar magnet with uniform magnetic feild inside

•Assumptions :

1) Magnetic field inside solenoid is uniform

B(inside) = Constant

2) Magnetic feild outside solenoid is very weak

B(outside) = 0

•Let the solenoid has N turns , If current 'i' starts in the solenoid it will result in uniform magnetic feild B(inside) in the solenoid .

•Let us consider rectangular loop PQRS as shown.

•Using Ampere's law

∫B.dl (PQRS) = μ•i(enclosed)

∫B.dl(PQ) + ∫B.dl(QR) + ∫B.dl(RS) = μ•i

+∫B.dl(SP)

•In QR & SP B is perpendicular to dl therefore B.dl =0

& In RS B(outside) =0 therefore

B.dl = 0

•So, ∫B.dl (PQ) = μ•i×No. of turns in

PQRSS loop

•since B(inside) is constant

B∫dl = μ•iN(l)/L

B.l = μ•iN(l)/L

B = μ•iN/L

•N/L = No. of turns per unit length therefore N/L = n (Given)

So, B = μ•ni

• Magnetic field due to long solenoid is B = μ•ni