The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through each turn of the coil is

Question

The inductance of a closed-packed coil of 400 turns is 8 mH. A current of 5 mA is passed through it. The magnetic flux through each turn of the coil is (A) (1/4 π) μ0 Wb (B) (1/2 π) μ0 Wb (C) (1/3 π) μ0 Wb (D) 0.4 μ0 Wb

Tardigrade Admin · Accepted Answer

N φ =L i ⇒ φ =(L i/N)=(8 × 10-3 × 5 × 10-3/400) =10-7=(μ0/4 π) Wb

Q. The inductance of a closed-packed coil of $400$ turns is $8 m H$ . A current of $5 m A$ is passed through it. The magnetic flux through each turn of the coil is

$\frac{1}{4 π} μ_{0} Wb$

$\frac{1}{2 π} μ_{0} Wb$

$\frac{1}{3 π} μ_{0} Wb$

$0.4 μ_{0} Wb$

$Nϕ = L i \Rightarrow ϕ$
$= \frac{L i}{N} = \frac{8 \times 1 0 ^{- 3} \times 5 \times 1 0 ^{- 3}}{400}$
$= 1 0^{- 7} = \frac{μ _{0}}{4 π} Wb$

Q. The inductance of a closed-packed coil of 400 turns is 8mH. A current of 5mA is passed through it. The magnetic flux through each turn of the coil is

4π1​μ0​Wb

2π1​μ0​Wb

3π1​μ0​Wb

0.4μ0​Wb