The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA is

Question

The self inductance of a coil having 400 turns is 10 mH. The magnetic flux through the cross section of the coil corresponding to current 2 mA is (A) 4× 10-5 Wb (B) 2× 10-3 Wb (C) 3× 10-5 Wb (D) 8× 10-3 Wb

Tardigrade Admin · Accepted Answer

The relation between the magnetic flux and the self-inductance of the coil is given by,

ϕ = N L I

Now, given that,
Number of turns

N = 400

Self-inductance

L = 10 m H = 10 \times 1 0^{- 3} H

Current,

I = 2 m A = 2 \times 1 0^{- 3} A

So, the total magnetic flux is given by,

ϕ = 400 (10 \times 1 0^{- 3}) \times 2 \times 1 0^{- 3}

\Rightarrow ϕ = 8 \times 1 0^{- 3} Wb

Q. The self inductance of a coil having $400$ turns is $10 m H$ . The magnetic flux through the cross section of the coil corresponding to current $2 m A$ is

$4 \times 1 0^{- 5} Wb$

$2 \times 1 0^{- 3} Wb$

$3 \times 1 0^{- 5} Wb$

$8 \times 1 0^{- 3} Wb$

Q. The self inductance of a coil having 400 turns is 10mH. The magnetic flux through the cross section of the coil corresponding to current 2mA is

4×10−5Wb

2×10−3Wb

3×10−5Wb

8×10−3Wb

The relation between the magnetic flux and the self-inductance of the coil is given by, ϕ=NLI Now, given that, Number of turns N=400 Self-inductance L=10mH=10×10−3H Current, I=2mA=2×10−3A So, the total magnetic flux is given by, ϕ=400(10×10−3)×2×10−3 ⇒ϕ=8×10−3Wb

Q. The self inductance of a coil having $400$ turns is $10 m H$ . The magnetic flux through the cross section of the coil corresponding to current $2 m A$ is

$4 \times 1 0^{- 5} Wb$

$2 \times 1 0^{- 3} Wb$

$3 \times 1 0^{- 5} Wb$

$8 \times 1 0^{- 3} Wb$