The self inductance of an inductor coil having 100 turns is 20 mH. The magnetic flux through the cross-section of the coil corresponding to a current of 4 mA is

Question

The self inductance of an inductor coil having 100 turns is 20 mH. The magnetic flux through the cross-section of the coil corresponding to a current of 4 mA is (A) 2× 10-5 Wb (B) 4× 10-7 Wb (C) 8× 10-7 Wb (D) 8× 10-5 Wb

Tardigrade Admin · Accepted Answer

Total magnetic flux, Nφ=LI=20×10-3×4×10-3 =8×10-5 Wb Magnetic flux through the cross-section of the coil, φ=(8×10-5/N)=(8×10-5/100) =8×10-7 Wb

Q. The self inductance of an inductor coil having $100$ turns is $20 m H$ . The magnetic flux through the cross-section of the coil corresponding to a current of $4 m A$ is

$2 \times 1 0^{- 5} Wb$

$4 \times 1 0^{- 7} Wb$

$8 \times 1 0^{- 7} Wb$

$8 \times 1 0^{- 5} Wb$

Total magnetic flux,
$Nϕ = L I = 20 \times 1 0^{- 3} \times 4 \times 1 0^{- 3}$
$= 8 \times 1 0^{- 5} Wb$
Magnetic flux through the cross-section of the coil,
$ϕ = \frac{8 \times 1 0 ^{- 5}}{N} = \frac{8 \times 1 0 ^{- 5}}{100}$
$= 8 \times 1 0^{- 7} Wb$

Q. The self inductance of an inductor coil having 100 turns is 20mH. The magnetic flux through the cross-section of the coil corresponding to a current of 4mA is

2×10−5Wb

4×10−7Wb

8×10−7Wb

8×10−5Wb